# Mole Concept & Stoichiometry Miscellaneous Exercises

## Miscellaneous Exercise

#### Question 1

From the equation for burning of hydrogen and oxygen

2H2 + O2 ⟶ 2H2O (Steam)

Write down the number of mole (or moles) of steam obtained from 0.5 moles of oxygen.

$\begin{matrix} 2\text{H}_2 & + &\text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2\text{ mole} & & 1\text{ mole}& & 2\text{ mole} \end{matrix}$

1 mole of oxygen gives 2 moles of steam

∴ 0.5 mole of oxygen will give $\dfrac{2}{1}$ × 0.5

= 1 mole of steam

#### Question 2

From the equation

3Cu + 8HNO3 ⟶ 3Cu(NO3)2 + 4H2O + 2NO

(At. mass Cu=64, H=1, N=14, O=16)

Calculate:

(a) Mass of copper needed to react with 63 g of HNO3

(b) Volume of nitric oxide at S.T.P. that can be collected.

$\begin{matrix} 3\text{Cu} & + &8\text{HNO}_3 & \longrightarrow & 3\text{Cu(NO}_3)_2 + 4\text{H}_2\text{O} + 2\text{NO} \\ 3(64) && 8(1 + 14 + 3(16)) \\ = 192 \text{g} & & = 8(63) \\ &&= 504 \text{g} \\ \end{matrix}$

(a) 504 g nitric acid reacts with 192 g of copper

∴ 63 g of nitric acid reacts with $\dfrac{192}{504}$ x 63 = 24 g of copper

Hence, 24 g of copper is required.

(b) 504 g of nitric acid gives 2 × 22.4 litre volume of NO

∴ 63 g of nitric acid gives $\dfrac{2 × 22.4}{504}$ x 63

= 5.6 litre of NO

5.6 L of NO is collected.

#### Question 3

(a) Calculate the number of moles in 7 g of nitrogen.

(b) What is the volume at S.T.P. of 7.1 g of chlorine?

(c) What is the mass of 56 cm3 of carbon monoxide at S.T.P?

(a) Gram molecular mass of N2 = 2 x 14 = 28 g

28 g of nitrogen = 1 mole

∴ 7 g of nitrogen = $\dfrac{1}{28}$ x 7

= 0.25 moles

(b) Gram molecular mass of Cl2 = 2 x 35.5 = 71 g

71 g of chlorine at S.T.P. occupies 22.4 litres

∴ 7.1 g of chlorine will occupy $\dfrac{22.4}{71}$ x 7.1

= 2.24 litre = 2.24 dm3

(c) Gram molecular mass of carbon monoxide (CO) = 12 + 16 = 28 g

28 g of carbon monoxide at S.T.P. occupies 22,400 cm3 volume

So, 22,400 cm3 volume have mass = 28 g

∴ 56 cm3 volume will have mass $\dfrac{28}{22,400}$ x 56 = 0.07 g

#### Question 4

Some of the fertilizers are sodium nitrate NaNO3, ammonium sulphate (NH4)2SO4 and urea CO(NH2)2. Which of these contains the highest percentage of nitrogen?

(i) Molar mass of NaNO3 = 23 + 14 + 3(16) = 23 + 14 + 48 = 85 g

Nitrogen in 85 g NaNO3 = 14 g

∴ Percentage of Nitrogen in NaNO3 = $\dfrac{14}{85}$ x 100 = 16.47%

(ii) Molar mass of (NH4)2SO4 = 2[14 + 4(1)] + 32 + 4(16) = 36 + 32 + 64 = 132 g

Nitrogen in 132 g of (NH4)2SO4 = 28 g

∴ Percentage of Nitrogen in (NH4)2SO4 = $\dfrac{28}{132}$ x 100 = 21.21%

(iii) Molar mass of CO(NH2)2 = 12 + 16 + 2[14 + (2 x 1)]

= 28 + 2(16)

= 28 + 32

= 60 g

Nitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g

∴ Percentage of Nitrogen in CO(NH2)2 = $\dfrac{28}{60}$ x 100 = 46.66%

So, the highest percentage of Nitrogen is in Urea.

#### Question 5

Water decomposes to O2 and H2 under suitable conditions as represented by the equation below:

2H2O ⟶ 2H2 + O2

(a) If 2500 cm3 of H2 is produced, what volume of O2 is liberated at the same time and under the same conditions of temperature and pressure?

(b) The 2500 cm3 of H2 is subjected to $2\dfrac{1}{2}$ times increase in pressure (temp. remaining constant). What volume will H2 now occupy?

(c) Taking the value of H2 calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 pressure remaining constant.

$\begin{matrix} 2\text{H}_2\text{O} & \longrightarrow & 2\text{H}_2 & + &\text{O} \\ 2\text{ Vol.} & & 2\text{ Vol.}& & 1\text{ Vol.} \end{matrix}$

2 Vol. of water gives 2 Vol. of H2 and 1 Vol. of O2

∴ If 2500 cm3 of H2 is produced, volume of O2 produced = $\dfrac{2500}{2}$ = 1250 cm3

(b) V1 = 2500 cm3

P1 = 1 atm = 760 mm

T1 = T

T2 = T

P2 = [760 x 2 $\dfrac{1}{2}$] + [760] = 760 [$\dfrac{5}{2}$ + 1] = 760 x $\dfrac{7}{2}$ = 2660 mm

V2 = ?

Using formula:

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

$\dfrac{760 \times 2500}{\text{T}}$ = $2660 \times \dfrac{\text{V}_2}{\text{T}}$

V2 = $\dfrac{760 \times 2500 }{2660}$ = $\dfrac{5000}{7}$

(c) V1 = $\dfrac{5000}{7}$ = 714.29 cm3

P1 = P2 = P

T1 = T

V2 = 2500 cm3

T2 = ?

Using formula:

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

$\dfrac{\text{P} \times 714.29}{\text{T}}$ = $\dfrac{\text{P} \times 2500}{\text{T}_2}$

T2 = $\dfrac{2500 }{714.29}$ x T

T2 = 3.5 x T

Hence, T2 = 3.5 times T or temperature should be increased by 3.5 times

#### Question 6

Urea (CO(NH2)2) is an important nitrogenous fertilizer. Urea is sold in 50 kg sacks. What mass of nitrogen is in one sack of urea?

Molar mass of urea [CO(NH2)2] = 12 + 16 + 2[14 + (2 x 1)]

= 28 + 2(16)

= 28 + 32

= 60 g

50 kg of urea = 50,000 g of urea

Nitrogen in 60 g of CO(NH2)2 = 2 x 14 = 28 g

∴ Nitrogen in 50,000 g urea = $\dfrac{28 \times 50,000 }{60}$

= 23,333 g = 23.3 kg

#### Question 7

Find the molecular formula of a hydrocarbon having vapour density 15, which contains 20% of Hydrogen.

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon8012$\dfrac{80}{12}$ = 6.66$\dfrac{6.66}{6.66}$ = 1
Hydrogen201$\dfrac{20}{1}$ = 20$\dfrac{20}{6.66}$ = 3

Simplest ratio of whole numbers = C : H = 1 : 3

Hence, empirical formula is CH3

Empirical formula weight = 12 + 3(1) = 15

V.D. = 15

Molecular weight = 2 x V.D. = 2 x 15

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 15}{15} = 2$

∴ Molecular formula = n[E.F.] = 2[CH3] = C2H6

#### Question 8

The following experiment was performed in order to determine the formula of a hydrocarbon. The hydrocarbon X is purified by fractional distillation.

0.145 g of X was heated with dry copper (II) oxide and 224 cm3 of carbon dioxide was collected at S.T.P.

(a) Which elements does X contain?

(b) What was the purpose of copper (II) oxide?

(c ) Calculate the empirical formula of X by the following steps:

(i) Calculate the number of moles of carbon dioxide gas.

(ii) Calculate the mass of carbon contained in this quantity of carbon dioxide and thus the mass of carbon in sample X.

(iii) Calculate the mass of hydrogen in sample X.

(iv) Deduce the ratio of atoms of each element in X (empirical formula).

(a) X contains carbon and hydrogen as it is a hydrocarbon.

(b) The purpose of copper (II) oxide is to act as an oxidizing agent.

(c) (i) 22400 cm3 CO2 has mass = 44 g

∴ 224 cm3 CO2 will have mass = $\dfrac{44}{22400}$ x 224 = 0.44 g

Molar mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g

Moles of CO2 = $\dfrac{\text{Mass}}{\text{{Molecular mass}}}$ = $\dfrac{0.44}{44}$ = 0.01 moles

(ii) Mass of carbon in 44 g CO2 = 12 g

∴ Mass of carbon in 0.44 g CO2 = $\dfrac{12}{44}$ x 0.44 = 0.12 g

As X contains carbon and hydrogen, so sample X has 0.12 g of carbon

(iii) Mass of Hydrogen in X = 0.145 - 0.12 = 0.025 g

(iv) Ratio of moles of C : H

= $\dfrac{0.01}{0.01}$ : $\dfrac{0.025}{0.01}$

= 1 : $\dfrac{25}{10}$

= 1 : $\dfrac{5}{2}$

= 2 : 5

Hence, ratio of C : H = 2 : 5

so, the empirical formula of hydrocarbon is C2H5

#### Question 9

Compound is formed by 24 g of X and 64 g of oxygen. If atomic mass of X = 12 and O = 16, calculate the simplest formula of compound.

ElementMass (g)Atomic massMolesSimplest ratio
Element X2412$\dfrac{24}{12}$ = 2$\dfrac{2}{2}$ = 1
Oxygen6416$\dfrac{64}{16}$ = 4$\dfrac{4}{2}$ = 2

Simplest ratio of whole numbers = X : O = 1 : 2

Hence, simplest formula of compound is XO2

#### Question 10

A gas cylinder filled with hydrogen holds 5 g of the gas. The same cylinder holds 85 g of gas X under the same temperature and pressure. Calculate:

(a) Vapour density of gas X.

(b) Molecular weight of gas X.

(a) V.D. = $\dfrac{\text{mass of certain volume of gas at S.T.P.}}{\text{mass of equal volume of H}_2\text{ at S.T.P.}}$ = $\dfrac{85}{5}$ = 17

(b) Molecular weight = 2 x V.D. = 2 x 17 = 34 a.m.u.

#### Question 11

(a) When carbon dioxide is passed over red hot carbon, carbon monoxide is produced according to the equation :

CO2 + C ⟶ 2CO

What volume of carbon monoxide at S.T.P. can be obtained from 3 g of carbon?

(b) 60 cm3 of oxygen was added to 24 cm3 of carbon monoxide and mixture ignited. Calculate:

(i) volume of oxygen used up and

(ii) Volume of carbon dioxide formed.

$\begin{matrix} \text{CO}_2 & + &\text{C} & \longrightarrow & 2\text{CO} \\ 1\text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

[By Gay Lussac's law]

1 Vol. of C produces 2V of CO

12 g of C produces 2 x 22.4 L of CO

∴ 3 g of C will produce $\dfrac{2 \times 22.4}{12}$ x 3

= 11.2 L of CO

(b)

$\begin{matrix} \text{O}_2 & + &\text{2CO} & \longrightarrow & 2\text{CO}_2 \\ 1\text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 2 \text{ vol.} \\ \end{matrix}$

(i) 2 Vol. of CO requires 1 Vol. of oxygen

∴ 24 cm3 CO will require $\dfrac{1}{2}$ x 24

= 12 cm3 of oxygen

(ii) 2 Vol. of CO gives 2 vol. of CO2

∴ 24 cm3 of CO will give 24 cm3 of CO2

#### Question 12

How much calcium oxide is obtained by heating 82 g of calcium nitrate? Also find the volume of NO2 evolved:

2Ca(NO3)2 ⟶ 2CaO + 4NO2 + O2

$\begin{matrix} 2\text{Ca(NO}_3)_2 & \longrightarrow & 2\text{CaO} & + &\space 4\text{NO}_2 \space + & \text{O}_2 \\ 2[40 + 2[14 + 3(16)]] & & 2[40 + 16] \\ = 2[40 + 28 + 96] & & = 2(56) \\ = 328 \text{ g} & & = 112 \text{ g} \end{matrix}$

328 g of calcium nitrate produces 112 g of CaO

∴ 82 g of calcium nitrate will produce $\dfrac{112}{328}$ x 82

= 28 g of CaO

328 g of calcium nitrate produces 4 x 22.4 L of NO2

∴ 82 g of calcium nitrate will produce $\dfrac{4 \times 22.4}{328}$ x 82

= 22.4 L of NO2

#### Question 13

The equation for the burning of octane is:

2C8H18 + 25O2 ⟶ 16CO2 + 18H2O

(i) How many moles of carbon dioxide are produced when one mole of octane burns?

(ii) What volume at S.T.P. is occupied by the number of moles determined in (i)?

(iii) If the relative molecular mass of carbon dioxide is 44, what is the mass of carbon dioxide produced by burning two moles of octane?

(iv) What is the empirical formula of octane?

$\begin{matrix} 2\text{C}_8\text{H}_{18} & + &25\text{O}_2 & \longrightarrow & 16\text{CO}_2 & + & 18\text{H}_2\text{O} \\ 2\text{ Vol.}&& 25\text{ Vol.} && 16\text{ Vol.}&& 18\text{ Vol.} \\ &&&&16[12 + 2(16)] \\ &&&&16[12 + 32] \\ &&&&16[44] = 704 \text{g} \\ \end{matrix}$

(i) 2 moles of octane produces 16 moles of CO2

∴ 1 mole octane produces $\dfrac{16}{2}$ x 1

= 8 moles of CO2

(ii) 1 mole CO2 occupies volume = 22.4 L

As 2 moles of octane produces 8 moles of carbon dioxide which will occupy volume $\dfrac{22.4}{1}$ x 8

= 179.2 dm3 of CO2

(iii) 1 mole CO2 has mass = 44 g

∴ 16 moles will have mass $\dfrac{44}{1}$ x 16

= 704 g of CO2

(iv) Molecular formula is C8H18

∴ Ratio of C and H is 8 : 18

Simple ratio is 4 : 9

Hence, empirical formula = C4H9

#### Question 14

Ordinary chlorine gas has two isotopes 3517Cl and 3717Cl in the ratio of 3:1. Calculate the relative atomic mass of chlorine.

The relative atomic mass of Cl = $\dfrac{(35 × 3) + (1 × 37)}{4}$ = 35.5

#### Question 15

Silicon (Si = 28) forms a compound with chlorine (Cl = 35.5) in which 5.6 g of silicon combines with 21.3 g of chlorine. Calculate the empirical formula of the compound.

ElementMassAt. wt.gram atomsSimplest ratio
Silicon5.628$\dfrac{5.6 }{28}$ = 0.2$\dfrac{0.2 }{0.2}$ = 1
Chlorine21.335.5$\dfrac{ 21.3}{35.5}$ = 0.6$\dfrac{ 0.6 }{0.2}$ = 3

Simplest ratio of whole numbers = Si : Cl = 1 : 3

Hence, empirical formula is SiCl3

#### Question 16

An acid of phosphorus has the following percentage composition; Phosphorus = 38.27%; hydrogen = 2.47%; oxygen = 59.26%. Find the empirical formula of the acid and it's molecular formula, given that it's relative molecular mass is 162.

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Phosphorus38.2731$\dfrac{38.27 }{31}$ = 1.234$\dfrac{ 1.234}{1.234 }$ = 1
Hydrogen2.471$\dfrac{2.47 }{1}$ = 2.47$\dfrac{2.47 }{1.234 }$ = 2
Oxygen59.2616$\dfrac{59.26}{16}$ = 3.70$\dfrac{3.70}{ 1.234 }$ = 3

Simplest ratio of whole numbers = P : H : O = 1 : 2 : 3

Hence, empirical formula is H2PO3

Empirical formula weight = 31 + 2(1) + 3(16) = 31 + 2 + 48 = 81

Molecular weight = 162

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{162}{81} = 2$

So, molecular formula = 2(H2PO3) = H4P2O6

#### Question 17

(a) Calculate the mass of substance 'A' which in gaseous form occupies 10 litres at 27°C and 700 mm pressure. The molecular mass of 'A' is 60.

(b) A gas occupied 360 cm3 at 87°C and 380 mm Hg pressure. If the mass of gas is 0.546 g, find it's relative molecular mass.

(a) Given, molecular mass of 'A' is 60

V1 = 10 L

T1 = 27 + 273 K = 300 K

P1 = 700 mm

T2 = 273 K

P2 = 760 mm

V2 = ?

Using formula:

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

Substituting in the formula,

$\dfrac{700 \times 10}{300}$ = $\dfrac{760 \times\text{V}_2}{273}$

V2 = $\dfrac{700 \times10 \times 273}{300 \times760}$ = $\dfrac{1911}{228}$ = 8.38 L

As, 22.4 L of A weighs 60 g

∴ 8.38 L of A will weigh $\dfrac{60}{22.4}$ x 8.38

= 22.446 = 22.45 g

(b) V1 = 360 cm3

T1 = 87 + 273 K = 360 K

P1 = 380 mm Hg pressure

T2 = 273 K

P2 = 760 mm Hg pressure

V2 = ?

Using formula:

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

Substituting in the formula,

$\dfrac{380\times 360}{360}$ = $\dfrac{760 \times\text{V}_2}{273}$

V2 = $\dfrac{380 \times 273}{760}$ = $\dfrac{10,374}{760}$ = 136.5 cm3

136.5 cm3 of gas weigh = 0.546 g

22400 cm3 of gas weigh $\dfrac{0.546}{136.5}$ x 22400

= 89.6 amu

#### Question 18

A gas cylinder can hold 1 kg of hydrogen at room temperature and pressure.

(a) What mass of carbon dioxide can it hold under similar conditions of temperature and pressure?

(b) If the number of molecules of hydrogen in the cylinder is X, calculate the number of carbon dioxide molecules in the cylinder. State the law that helped you to arrive at the above result

(a) Given, cylinder can hold = 1 kg of hydrogen

Molar mass of hydrogen = 2[1] = 2g

Number of moles of hydrogen present in cylinder = $\dfrac{1000}{2}$ = 500

According to avogadro's law : cylinder will hold 500 moles of carbon dioxide gas

Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g

1 mole of carbon dioxide = 44 g

∴ 500 moles of carbon dioxide = 44 x 500 = 22,000 g = 22 kg.

Hence, Weight of carbon dioxide in cylinder = 22 kg

(b) According to Avogadro's law — Equal volumes of all gases under similar conditions of temperature and pressure contain equal number of molecules.

∴ molecules in the cylinder of carbon dioxide = X

Avogadro's law helped to arrive at this result.

#### Question 19

Following questions refer to one mole of chlorine gas.

(a) What is the volume occupied by this gas at S.T.P.?

(b) What will happen to the volume of gas, if pressure is doubled?

(c) What volume will it occupy at 273°C?

(d) If the relative atomic mass of chlorine is 35.5, what will be the mass of 1 mole of chlorine gas?

(a) According to Avogadro's law : The volume occupied by 1 mole of chlorine is 22.4 dm3

(b) According to Boyle's Law: PV = constant

Hence, if pressure is doubled then volume will become half i.e. $\dfrac{22.4}{2}$ = 11.2 dm3

(c) V1 = 22.4 dm3

T1 = 273 K

T2 = 273 + 273 K = 546 K

V2 = ?

According to charles law:

$\dfrac{\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{V}_2}{\text{T}_2}$

Substituting to we get,

$\dfrac{ 22.4}{273}$ = $\dfrac{\text{V}_2}{546}$

Hence, V2 = $\dfrac{ 22.4}{273}$ x 546 = 44.8 dm3

(c) Mass of 1 mole Cl2 gas = 35.5 × 2 = 71 g

#### Question 20

(a) A hydrate of calcium sulphate CaSO4.xH2O contains 21% water of crystallisation. Find the value of x.

(b) What volume of hydrogen and oxygen measured at S.T.P. will be required to prepare 1.8 g of water.

(c) How much volume will be occupied by 2g of dry oxygen at 27°C and 740 mm pressure?

(d) What would be the mass of CO2 occupying a volume of 44 litres at 25°C and 750 mm pressure?

(e) 1 g of a mixture of sodium chloride and sodium nitrate is dissolved in water. On adding silver nitrate solution, 1.435 g of AgCl is precipitated.

AgNO3 (aq) + NaCl (aq) ⟶ AgCl (s) + NaNO3

Calculate the percentage of NaCl in the mixture.

Relative molecular mass of CuSO4.xH2O

= 40 + 32 + (4×16) + [x(2+16)]

= 40 + 32 + 64 + 18x

= 136 + 18x

∴ 21% water of crystallization = 18 x

$\Rightarrow \dfrac{18x}{136 + 18x } = \dfrac{21}{100} \\[1em] 1800x = 21(136 + 18x) \\[1em] 1800x = 2856 + 378x \\[1em] 1800x - 378x = 2856 \\[1em] 1422x = 2856 \\[1em] x = \dfrac{2856}{1422} \\[1em] x = 2$

Hence, water of crystallization = 2

(b) Molar mass of H2O = 2(1) + 16 = 18 g

For 18 g water, vol. of hydrogen needed = 22.4 litre

∴ For 1.8 g, vol. of H2 needed = $\dfrac{22.4}{18}$ x 1.8 = 2.24 L

According to Gay Lussac's Law, 2 vol of hydrogen requires 1 vol of oxygen

When 2 vol of hydrogen in 1.8 g H2O is 2.24 L, then one vol. of oxygen will be:

$\dfrac{2.24}{2}$ = 1.12 L

(c) Gram molecular mass of O2 = 2 x 16 = 32 g

1 mole of O2 weighs 32 g and occupies 22.4 lit. vol.

∴ 2 g of O2 occupies = $\dfrac{22.4}{32} \times 2 = 1.4 \text{ lit.}$

Volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure:

S.T.P.Given Values
P1 = 760 mm of HgP2 = 740 mm of Hg
V1 = 1.4 litV2 = x lit
T1 = 273 KT2 = 27 + 273 K

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{760 \times 1.4}{273} = \dfrac{740 \times x}{300} \\[0.5em] x = \dfrac{760 \times 1.4 \times 300}{740 \times 273 } \\[0.5em] x = \dfrac{3,19,200}{2,02,020} \\ \\[0.5em] x = 1.58 \text{ lit}$

Hence, the volume occupied by 2 g of O2 gas at 27°C and 740 mm pressure is 1.58 lit.

(d) Gram molecular mass of CO2 = 12 + 2(16) = 12 + 32 = 44 g

Given ValuesS.T.P.
P1 = 750 mm of HgP2 = 760 mm of Hg
V1 = 44 litV2 = x lit
T1 = 25 + 273 K = 298 KT2 = 273 K

Using the gas equation,

$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$

Substituting the values we get,

$\dfrac{750 \times 44}{298} = \dfrac{760 \times x}{273} \\[0.5em] x = \dfrac{750 \times 44 \times 273}{760 \times 298 } \\[0.5em] x = \dfrac{9,009,000}{226,480} \\ \\[0.5em] x = 39.78 \text{ lit}$

22.4 litre of CO2 at S.T.P. has mass = 44 g

39.78 litre of CO2 at S.T.P. has mass $\dfrac{44}{22.4}$ x 39.78

= 78.14 g

(e)

$\begin{matrix} \text{AgNO}_3 & + &\text{NaCl} & \longrightarrow & \text{AgCl} & + & \text{NaNO}_3 \\ && 23 + 35.5 && 108 + 35.5 \\ && = 58.5 \text{g}&& = 143.5\text{g} \\ \end{matrix}$

(i) 143.5 g AgCl is formed by 58.5 g NaCl

∴ 1.435 g of AgCl will be formed by $\dfrac{58.5}{143.5}$ x 1.435 = 0.582 g

Percentage of NaCl = $\dfrac{0.582}{1}$ x 100 = 58.5%

Hence, percentage of NaCl is 58.5%

#### Question 21a

From the equation:

C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2

Calculate:

(i) The mass of carbon oxidized by 49 g of sulphuric acid.

(ii) The volume of sulphur dioxide measured at STP, liberated at the same time.

$\begin{matrix} \text{C} & + &2\text{H}_2\text{SO}_4 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O}& + & 2\text{SO}_2 \\ 12 \text{g} && 2[2(1) + 32 + 4(16)] \\ && 2[2 + 32 + 64] \\ && 196 \text{g} \\ \end{matrix}$

(i) 196 g of sulphuric acid oxidizes 12 g carbon

∴ 49 g of sulphuric acid will oxidize = $\dfrac{12}{196}$ x 49 = 3 g

Hence, 3 g of carbon is oxidized.

(ii) 12 g carbon liberates 2 vol = (2 x 22.4) lit of SO2

∴ 3 g of carbon will liberate $\dfrac{2 \times 22.4}{12}$ x 3 = 11.2 lit of SO2.

Hence, 11.2 lit of SO2 is liberated.

#### Question 21b

(i) A compound has the following percentage composition by mass: carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. (H = 1; C = 12; Cl = 35.5)

(ii) The relative molecular mass of this compound is 168, so what is it's molecular formula?

(i)

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Carbon14.412$\dfrac{14.4}{12}$ = 1.2$\dfrac{1.2}{1.2}$ = 1
Hydrogen1.21$\dfrac{1.2}{1}$ = 1.2$\dfrac{1.2}{1.2}$ = 1
chlorine84.535.5$\dfrac{84.5}{35.5}$ = 2.38$\dfrac{2.38}{1.2}$ = 1.98 = 2

Simplest ratio of whole numbers = C : H : Cl = 1 : 1 : 2

Hence, empirical formula is CHCl2

Empirical formula weight = 12 + 1 + 2(35.5) = 84 g

Relative molecular mass = 168

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{168}{84} = 2$

Molecular formula = n[E.F.] = 2[CHCl2] = C2H2Cl4

Molecular formula = C2H2Cl4

#### Question 22

Find the percentage of

(a) oxygen in magnesium nitrate crystals [Mg(NO3)2.6H2O].

(b) boron in Na2B4O7.10H2O. H=1,B=11,O=16,Na=23.

(c) phosphorus in the fertilizer superphosphate Ca(H2PO4)2

Relative molecular mass of [Mg (NO3)2.6H2O]

= 24 + 2[14 + 3(16)] + 12(1) + 6(16)
= 24 + 2[14 + 48] + 12 + 96
= 24 + 28 + 96 + 12 + 96
= 256 g

Molar mass of oxygen in [Mg (NO3)2.6H2O] = 96 + 96 = 192 g

Since, 256 g of [Mg(NO3)2.6H2O] contains 192 g of oxygen

∴ 100 g of [Mg(NO3)2.6H2O] contains

$\dfrac{192}{256}$ x 100 = 75% of oxygen

(b) Relative molecular mass of Na2B4O7.10H2O
= 2(23) + 4(11) + 7(16) + 20(1) + 10(16)
= 46 + 44 + 112 + 20 + 160
= 382 g

Molar mass of boron in Na2B4O7.10H2O = 44 g

382 g of Na2B4O7.10H2O contains 44 g of boron

∴ 100 g Na2B4O7.10H2O contains $\dfrac{44}{382}$ x 100 = 11.5%

(c) Relative molecular mass of Ca(H2PO4)2
= 40 + 4(1) + 2(31) + 8(16)
= 40 + 4 + 62 + 128
= 234 g

Molar mass of phosphorus in Ca(H2PO4)2 = 62 g

234g of Ca(H2PO4)2 contains 62 g of phosphorus

∴ 100 g of Ca(H2PO4)2 contains $\dfrac{62}{234}$ x 100 = 26.5%

#### Question 23

What mass of copper hydroxide is precipitated by using 200 g of sodium hydroxide?

2NaOH + CuSO4 ⟶ Na2SO4 + Cu(OH)2

[Cu = 64, Na = 23, S = 32, H = 1]

$\begin{matrix} 2\text{NaOH} & + & \text{CuSO}_4 & \longrightarrow & \text{Na}_2\text{SO}_4 & + & \text{Cu(OH)}_2\downarrow \\ 2[23 + 16 & & & & & & 64 + 2[16 + 1] \\ + 1] & & & & & & = 64 + 34 \\ = 80 \text{ g} & & & & & & = 98 \text{g} \\ \end{matrix}$

80 g of sodium hydroxide precipitates 98 g of copper hydroxide.

∴ 200 g of sodium hydroxide will precipitate $\dfrac{98}{80}$ x 200 = 245 g of copper hydroxide.

Hence, 245 g. of copper hydroxide is precipitated.

#### Question 24

Solid ammonium dichromate decomposes as under:

(NH₄)₂Cr₂O₇ ⟶ N₂ + Cr₂O₃ + 4H₂O

If 63 g of ammonium dichromate decomposes. Calculate

(a) the quantity in moles of (NH₄)₂Cr₂O₇

(b) the quantity in moles of nitrogen formed.

(c) the volume of N₂ evolved at S.T.P.

(d) the loss of mass

(iv) the mass of chromium (III) formed at the same time.

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & \xrightarrow{\Delta} & \text{N}_2 & + & 4\text{H}_2\text{O} & + & \text{Cr}_2\text{O}_3 \\ 2[14 + 4(1)] & & & & & & 2(52) \\ + 2(52) + 7(16) & & & & & & + 3(16) \\ = 36 + 104 & & & & & & = 104 + 48 \\ + 112 = 252 \text{ g} & & & & & & = 152 \text{ g} \\ \end{matrix}$

(a) 252 g of (NH4)2Cr2O7 = 1 mole

∴ 63 g of (NH4)2Cr2O7 = $\dfrac{1}{252}$ x 63 = 0.25 moles

Hence, no. of moles = 0.25 moles

(b)

$\begin{matrix} (\text{NH}_4)_2\text{Cr}_2\text{O}_7 & : & \text{N}_2 \\ 1 \text{ mol.} & : & 1 \text{ mol.} \\ 0.25 \text{ mol.} & : & x \\ \end{matrix}$

Hence, 0.25 moles of (NH4)2Cr2O7 will produce 0.25 moles of nitrogen.

(c) 1 mole of N₂ occupies 22.4 lit.

∴ 0.25 moles of N₂ will occupy = 22.4 x 0.25 = 5.6 lit.

Hence, volume of N₂ evolved at s.t.p = 5.6 lit.

(d) 252 g of (NH4)2Cr2O7 decomposes to give 152 g of solid Cr₂O₃, loss in mass = 252 - 152 = 100 g

If 63 g of (NH4)2Cr2O7 decomposes then loss in mass is

$\dfrac{100}{252}$ x 63 = 25 g

(e) 1 mole of Cr2O7 = 152 g.

∴ 0.25 moles of Cr2O7 = 152 x 0.25 = 38 g.

Hence, mass in gms of Cr2O7 formed = 38 g.

#### Question 25

Hydrogen sulphide gas burns in oxygen to yield 12.8 g of sulphur dioxide gas as under:

2H2S + 3O2 ⟶ 2H2O + 2SO2

Calculate the volume of hydrogen sulphide at S.T.P. Also, calculate the volume of oxygen required at S.T.P. which will complete the combustion of hydrogen sulphide determined in (litres).

$\begin{matrix} 2\text{H}_2\text{S} & + &3\text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} & + & 2\text{SO}_2 \\ 2\text{ Vol.} && 3\text{ Vol.} && && 2\text{ Vol.} \\ 2[2(1) + 32] && 3[2(16)] &&&& 2[32 + 2(16)] \\ 68 \text{ g} && 96 \text{ g} &&&& 128 \text{ g} \end{matrix}$

128 g of SO2 has volume 2 × 22.4 litres

∴ 12.8 g of SO2 has volume =

$\dfrac{2 × 22.4}{128}$ x 12.8 = 4.48 L

Volume of oxygen = ?

2 × 22.4 L H2S requires = 3 × 22.4 litre of oxygen

∴ 4.48 L H2S will require

$\dfrac{3 × 22.4}{2 × 22.4 }$ x 4.48 = 6.72 L of oxygen.

#### Question 26

Ammonia burns in oxygen and combustion, in the presence of a catalyst, may be represented by

2NH3 + 2 $\dfrac{1}{2}$ O2 ⟶ 2NO + 3H2O

[H = 1, N = 14, O = 16]

What mass of steam is produced when 1.5 g of nitrogen monoxide is formed?

$\begin{matrix} 2\text{NH}_3 & + &2\dfrac{1}{2}\text{O}_2 & \longrightarrow & 2\text{NO} & + &3\text{H}_2\text{O} \\ &&&&2[14 + 16] && 3[2(1) + 16] \\ &&&& 60 \text{ g} && 54 \text{ g} \\ \end{matrix}$

When 60 g NO is formed then, 54 g mass of steam is produced

∴ when 1.5 g NO is formed, mass of steam produced

= $\dfrac{54}{60}$ x 1.5 = 1.35 g

#### Question 27

If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer, calcium nitrate Ca(NO3)2 would be required to replace the nitrogen in a 10 hectare field ?

Molecular mass of Ca(NO3)2

= 40 + [2(14) + 6(16)]
= 40 + 28 + 96
= 164 g

Mass of N2 in Ca(NO3)2 = 28 g

In 1 hectare of soil, 20 kg of N2 is removed

∴ In 10 hectare field N2 removed is 20 x 10 = 200 kg

28 g N2 is present in 164 g of Ca(NO3)2

So, 200 kg or 200,000 g of N2 is present in $\dfrac{164}{28}$ x 200,000

= 1171428 g = 1171.4 kg

#### Question 28

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5HNO3 ⟶ H3PO4+ 5NO2 + H2O

If 6.2 g of phosphorus was used in the reaction calculate:

(a) Number of moles of phosphorus taken and mass of phosphoric acid formed.

(b) mass of nitric acid consumed at the same time?

(c) The volume of steam produced at the same time if measured at 760 mm Hg pressure and 273°C?

$\begin{matrix} \text{P}& + &5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & 5\text{NO}_2& + &\text{H}_2\text{O} \\ 31 \text{g} && 5[1 + 14 + 3(16)] && 3(1) + 31 + 4(16) \\ && 315 \text{ g} && 98 \text{ g} \end{matrix}$

(a) 31 g of P = 1 mole

∴ 6.2 g of P = $\dfrac{1}{31}$ x 6.2 = 0.2 mole of P

Mass of phosphoric acid formed = ?

31 g of P produces 98 g of phosphoric acid

∴ 6.2 g of P will form $\dfrac{98}{31}$ x 6.2 = 19.6 g

(b) 31 g P reacts with 315 g HNO3

∴ 6.2 g P will react with $\dfrac{315}{31}$ x 6.2 = 63 g HNO3

(c) 31 g P produces = 1 mole steam

∴ 6.2 g P produces $\dfrac{1}{31}$ x 6.2 = 0.2 moles

Volume of steam produced at STP = 0.2 × 22.4 = 4.48 litre

V1 = 4.48 litre

T1 = 273 K

P1 = 760 mm Hg pressure

T2 = 273 + 273 = 546 K

P2 = 760 mm Hg pressure

V2 = ?

Using formula:

$\dfrac{\text{P}_1\text{V}_1}{\text{T}_1}$ = $\dfrac{\text{P}_2\text{V}_2}{\text{T}_2}$

Substituting in the formula,

$\dfrac{760\times 4.48}{273}$ = $\dfrac{760 \times\text{V}_2}{546}$

V2 = 2 x 4.48 = 8.96 L

Hence, volume of steam produced = 8.96 L

#### Question 29

112 cm3 of a gaseous fluoride of phosphorus has a mass of 0.63 g. Calculate the relative molecular mass of the fluoride. If the molecule of the fluoride contains only one atom of phosphorus, then determine the formula of the phosphorus fluoride. [ F = 19, P = 31]

Given,

112 cm3 of gaseous fluoride has mass = 0.63 g

so, 22400 cm3 of gaseous fluoride will have mass = $\dfrac{0.63}{112}$ x 22400

= 126 g

Relative molecular mass of fluoride = 126 g

The molecular mass = At mass P + At. mass of F

126 = 31 + At. Mass of F

∴ At. Mass of F = 126 - 31 = 95 g

However, At. mass of F = 19

$\dfrac{95}{19}$ = 5

So, 5 atoms of F, hence, the molecular formula = PF5

#### Question 30

Washing soda has formula Na2CO3.10H2O. What mass of anhydrous sodium carbonate is left when all the water of crystallization is expelled by heating 57.2 g of washing soda?

$\begin{matrix} \text{Na}_2\text{CO}_3.10\text{H}_2\text{O} & \xrightarrow{\Delta} & \text{Na}_2\text{CO}_3 & + & 10\text{H}_2\text{O} \\ 2(23) + 12 + 3 (16) & & 2(23) + 12 & & 10[2(1) \\ + 10[2(1) + 16] & & + 3(16) & & + 16] \\ = 46 + 12 + 48 & & = 46 + 12 & & = 180 \text{ g} \\ + 10(18) & & + 48 \\ = 286 \text{ g} & & = 106 \text{ g} \\ \end{matrix}$

286 g of washing soda had 106 g of anhydrous sodium carbonate

∴ 57.2 g will have = $\dfrac{106}{286}$ x 57.2 = 21.2 g anhydrous sodium carbonate.

Hence, 21.2 g anhydrous sodium carbonate is left.

#### Question 31

A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride (M = 56).

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Metal M34.556$\dfrac{34.5}{56}$ = 0.616$\dfrac{0.616}{0.616}$ = 1
chlorine65.535.5$\dfrac{65.5}{35.5}$ = 1.84$\dfrac{1.84}{0.616}$ = 2.98 = 3

Simplest ratio of whole numbers = M : Cl = 1 : 3

Hence, empirical formula is MCl3

Empirical formula weight = 56 + 3(35.5) = 162.5 g

Molecular weight = 2 x V.D. = 2 x 162.5 = 325

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{ 325}{162.5} = 2$

Molecular formula = n[E.F.] = 2[MCl3] = M2Cl6

Molecular formula = M2Cl6

#### Question 32

A compound X consists of 4.8% carbon and 95.2% bromine by mass.

(i) Determine the empirical formula of this compound working correct to one decimal place (C = 12; Br = 80)

(ii) If the vapour density of the compound is 252, what is the molecular formula of the compound?

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
carbon4.812$\dfrac{4.8}{12}$ = 0.4$\dfrac{0.4}{0.4}$ = 1
bromine95.280$\dfrac{95.2}{80}$ = 1.19$\dfrac{1.19}{0.4}$ = 2.9 = 3

Simplest ratio of whole numbers = C : Br = 1 : 3

Hence, empirical formula is CBr3

(ii) Empirical formula weight = 12 + 3(80) = 252 g

Molecular weight = 2 x V.D. = 2 x 252 = 504

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{ 504}{252} = 2$

Molecular formula = n[E.F.] = 2[CBr3] = C2Br6

Molecular formula = C2Br6

#### Question 33

The reaction: 4N2O + CH4 ⟶ CO2 + 2H2O + 4N2 takes place in the gaseous state. If all volumes are measured at the same temperature and pressure, calculate the volume of dinitrogen oxide (N2O) required to give 150 cm3 of steam.

$\begin{matrix} 4\text{N}_2\text{O} & + &\text{CH}_4 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} & + &4\text{N}_2 \\ 4 \text{ Vol} && 1 \text{ Vol} && 1 \text{ Vol} && 2 \text{ Vol} && 4\text{ Vol} \\ \end{matrix}$

2 x 22400 litre steam is produced by 4 × 22400 cm3 N2O

∴ 150 cm3 steam will be produced by

= $\dfrac{4 × 22400}{2 × 22400}$ x 150 = 300 cc of N2O

#### Question 34

Samples of the gases O2, N2, CO2 and CO under the same conditions of temperature and pressure contain the same number of molecules x. The molecules of oxygen occupy V litres and have a mass of 8 g under the same conditions of temperature and pressure

What is the volume occupied by:

(a) x molecules of N2

(b) 3x molecules of CO

(c) What is the mass of CO2 in grams?

(d) In answering the above questions, which law have you used?

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

∴ If gases under the same conditions have same number of molecules then they must have the same volume.

(i) So, X molecules of N2 occupy 1V litres.

(ii) 3X molecules of CO will occupies 3V litres.

(iii) Molar mass of CO2 = C + 2(O) = 12 + 2(16) = 44 g

1 mole of O2 weighs 32 g and occupy 22.4 lit. volume

∴ 8 g of O2 will occupy $\dfrac{22.4}{32}$ x 8 = 5.6 lit. vol. = Molar volume

If 8 g of O2 occupies 5.6 lit. vol.

And 1 mole of CO2 occupy 22.4 lit and weighs = 44 g at s.t.p.

∴ 5.6 lit. vol. of CO2 will weigh = $\dfrac{44}{22.4}$ x 5.6 = 11 g

Hence, mass of CO2 = 11 g

(iv) Avogadro's Law is used above.

#### Question 35

The percentage composition of sodium phosphate as determined by analysis is 42.1% sodium, 18.9% phosphorus and 39% oxygen. Find the empirical formula of the compound?

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
sodium42.123$\dfrac{42.1}{23}$ = 1.83$\dfrac{1.83}{0.609}$ = 3
phosphorus18.931$\dfrac{18.9}{31}$ = 0.609$\dfrac{0.609}{0.609}$ = 1
oxygen3916$\dfrac{39}{16}$ = 2.43$\dfrac{2.43}{0.609}$ = 4

Simplest ratio of whole numbers = Na : P : O = 3 : 1 : 4

Hence, empirical formula is Na3PO4

#### Question 36

What volume of oxygen is required to burn completely a mixture of 22.4 dm3 of methane and 11.2 dm3 of hydrogen into carbon dioxide and steam?

CH4 + 2O2 ⟶ CO2 + 2H2O

2H2 + O2 ⟶ 2H2O

$\begin{matrix} \text{CH}_4 & + & 2\text{O}_2 & \longrightarrow & \text{CO}_2 & + & 2\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 2 \text{ vol.} & \longrightarrow & 1\text{ vol.} \\ \end{matrix}$

From equation:

22.4 dm3 of methane requires oxygen = 2 x 22.4 dm3 of O2 = 44.8 dm3

$\begin{matrix} 2\text{H}_2 & + & \text{O}_2 & \longrightarrow & 2\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \\ \end{matrix}$

From equation,

[2 x 22.4] dm3 hydrogen requires oxygen = 22.4 dm3

∴ 11.2 dm3 hydrogen will require oxygen = $\dfrac{22.4}{2 \times 22.4}$ x 11.2 =

= 5.6 dm3

Total volume of oxygen required = 44.8 + 5.6 = 50.4 dm3

#### Question 37

The gases hydrogen, oxygen, carbon dioxide, sulphur dioxide and chlorine are arranged in order of their increasing relative molecular masses. Given 8 g of each gas at S.T.P., which gas will contain the least number of molecules and which gas the most?

Equal volumes of all gases, under similar conditions of temperature and pressure, contain equal numbers of molecules.

So, 1 mole of each gas contains = 6.02 × 1023 molecules

Mol. Mass of :

H2 = 2 g,

O2 = 32 g,

CO2 = 12 + 2(16) = 44 g,

SO2 = 32 + 2(16) = 64 g,

Cl2 = 2(35.5) = 71 g

(i) 2 g of hydrogen contains molecules = 6.02 × 1023

So, 8 g of hydrogen contains molecules

= $\dfrac{6.02 × 10^{23}}{2}$ x 8

= 4 × 6.02 × 1023molecules

= 24.08 × 1023molecules

(ii) 32 g of oxygen contains molecules = 6.02 × 1023

So, 8 g of oxygen contains molecules

= $\dfrac{6.02 × 10^{23}}{32}$ x 8

= 1.505 × 1023molecules

(iii) 44 g of carbon dioxide contains molecules = 6.02 × 1023

So, 8 g of carbon dioxide contains

= $\dfrac{6.02 × 10^{23}}{44}$ x 8

= 1.09 × 1023molecules

(iv) 64 g of sulphur dioxide contains molecules = 6.02 × 1023

So, 8g of sulphur dioxide contains

= $\dfrac{6.02 × 10^{23}}{64}$ x 8

= 0.75 × 1023molecules

(v) 71 g of chlorine contains molecules = 6.02 × 1023

So, 8g of chlorine contains

= $\dfrac{6.02 × 10^{23}}{71}$ x 8

= 0.67 × 1023molecules

Thus Cl2 < SO2 < CO2 < O2 < H2

(i) Least number of molecules in Cl2

(ii) Most number of molecules in H2

#### Question 38

10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation given below:

Na2SO4 + BaCl2 ⟶ BaSO4 + 2NaCl

Calculate the percentage of sodium sulphate in the original mixture.

$\begin{matrix} \text{Na}_2\text{SO}_4 & + & \text{BaCl}_2 & \longrightarrow & \text{BaSO}_4 \downarrow & + & 2\text{NaCl} \\ 2(23) + 32 & & & & 137 + 32 \\ + 4(16) & & & & + 4(16) \\ = 46 + 32 & & & & = 137 + 32 \\ + 64 & & & & + 64 \\ = 142 \text{ g} & & & & = 233 \text{ g} \\ \end{matrix}$

233 g of BaSO4 is obtained from 142 g of Na2SO4

6.99 g of BaSO4 will be obtained from $\dfrac{142}{233}$ x 6.99 = 4.26 g of Na2SO4.

∴ In 10 g mixture $\dfrac{4.26}{10}$ x 100 = 42.6% Na2SO4 is present.

Hence, 42.6% Na2SO4 is present in the original mixture

#### Question 39

When heated, potassium permanganate decomposes according to the following equation:

2KMnO4 ⟶ K2MnO4 + MnO2 + O2

(a) Some potassium permanganate was heated in the test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.

(b) Given that the molecular mass of potassium permanganate is 158. What volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres)

2KMnO4 ⟶ K2MnO4 + MnO2 + O2

Loss in mass = 1.32 g = 1 lit of oxygen

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{1.32}{0.0825} \\[0.5em] = 16 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 16 = 32 g

Hence, relative molecular mass of oxygen is 32 g

(b) Molar mass of 2KMnO4 = 2[39 + 55 + 4(16)] = 2[39 + 55 + 64] = 316 g

316 g of KMnO4 gives oxygen = 24 litres

∴ 15.8 g of KMnO4 will give

= $\dfrac{24}{316}$ x 15.8 = 1.2 L

#### Question 40

(a) A flask contains 3.2 g of sulphur dioxide. Calculate the following:

(i) The moles of sulphur dioxide present in the flask.

(ii) The number of molecules of sulphur dioxide present in the flask.

(iii) The volume occupied by 3.2 g of sulphur dioxide at S.T.P.

(b) An Experiment showed that in a lead chloride solution, 6.21 g of lead is combined with 4.26 g of chlorine. What is the empirical formula of this chlorine? (Pb = 207; Cl = 35.5)

(a) (i) Molar mass of sulphur dioxide = 32 + 2(16) = 64 g

64 g of sulphur dioxide = 1 mole

So, 3.2 g = $\dfrac{1}{64}$ x 3.2 = 0.05 moles

(ii) 1 mole of SO2 = 6.02 × 1023 molecules

So, in 0.05 moles, no. of molecules = 6.02 × 1023 × 0.05 = 3 × 1022

(iii) The volume occupied by 64 g of SO2 = 22.4 dm3

3.2 g of SO2 will be occupied by volume

$\dfrac{22.4}{64}$ x 3.2 = 1.12 L

(b) ElementMass (g)Atomic massMolesSimplest ratio
Pb6.21207$\dfrac{6.21}{207}$ = 0.03$\dfrac{0.03}{0.03}$ = 1
Cl4.2635.5$\dfrac{4.26}{35.5 }$ = 0.12$\dfrac{0.12}{0.03}$ = 4

Simplest ratio of whole numbers Pb : Cl = 1 : 4

Hence, empirical formula is PbCl4

#### Question 41

The volume of gases A, B, C and D are in the ratio, 1 : 2 : 2 : 4 under the same conditions of temperature and pressure .

(i) Which sample of gas contains the maximum number of molecules?

(ii) If the temperature and pressure of gas A are kept constant, then what will happen to the volume of A when the number of molecules is doubled?

(iii) If this ratio of gas volume refers to the reactants and products of a reaction, which gas law is being observed?

(iv) If the volume of A is actually 5.6 dm3 at S.T.P., calculate the number of molecules in the actual Volume of D at S.T.P. (Avogadro's number is 6 × 1023).

(v) Using your answer from (iv), state the mass of D if the gas is dinitrogen oxide (N2O)

(i) Volume is directly proportional to the number of molecules, hence gas D will have maximum no. of molecules as its volume is maximum.

(ii) If number of molecules of gas A is doubled, the volume will also be doubled i.e. 2V.

(iii) Gay Lussac's Law is observed.

(iv) 1 mole contains 6 x 1023 number of molecules and occupies 22.4 lit. vol.

Given, volume of 'A' is 5.6 dm3 at s.t.p.

∴ vol. of D will be 4 × 5.6 = 22.4 lit.
No. of molecules in 22.4 lit. of D = 6 x 1023 (Avogadro no.)

(v) As D is 1 mole hence, mass of 1 mole of D (N2O) = 2(14) + 16 = 28 + 16 = 44 g

Hence, mass of N2O = 44 g

#### Question 42

The equations given below relate to the manufacture of sodium carbonate [Mol. wt. of Na2CO3 = 106]

(i) NaCl + NH3 + CO2 + H2O ⟶ NaHCO3 + NH4Cl

(ii) 2NaHCO3 ⟶ Na2CO3 + H2O + CO2

Equations (1) and (2) are based on the production of 21.2 g. of sodium carbonate.

(a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g. of sodium carbonate

(b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required.

$\begin{matrix} 2\text{NaHCO}_3 & \longrightarrow & \text{Na}_2\text{CO}_3 & + & \text{H}_2\text{O} + \text{CO}_2 \\ 2 \text{mol.} & & 1 \text{mol.} \\ 2 \times 84 & & 106\text{ g.} \\ = 168\text{ g.} & & \end{matrix}$

106 g of Na2CO3 is obtained from 168 g of NaHCO3

∴ 21.2 g. of Na2CO3 is obtained from $\dfrac{168}{106}$ x 21.2 = 33.6 g of NaHCO3

(ii)

$\begin{matrix} \text{NaCl} & + & \text{NH}_3 & + & \text{CO}_2 & + & \text{H}_2\text{O} \\ & & & & 1\text{mol.} & \\ & & & & 22.4 \text{lit.} \\ \longrightarrow & \text{NaHCO}_3 & + & \text{NH}_4\text{Cl} \\ & 1 \text{mol.} \\ & 84\text{ g.} \end{matrix}$

84 g of NaHCO3 is obtained from 22.4 lit of CO2

∴ 33.6 g. of NaHCO3 is obtained from $\dfrac{22.4}{84}$ x 33.6 = 8.96 lit.

Hence, 8.96 lit of CO2 is required.

#### Question 43

A sample of ammonium nitrate when heated yields 8.96 litres of steam (measured at STP)

NH4NO3 ⟶ N2O + 2H2O

(i) What volume of di nitrogen oxide is produced at the same time as 8.96 litres of steam.

(ii) What mass of ammonium nitrate should be heated to produce 8.96 litres of steam [Relative molecular mass of NH4NO3 is 80]

(iii) Determine the percentage of oxygen in ammonium nitrate [O = 16]

$\begin{matrix} \text{NH}_4\text{NO}_3 &\xrightarrow{\Delta} & \text{N}_2\text{O} & + & 2\text{H}_2\text{O} \\ 1\text{ vol.} = 80 \text{g} & & 1 \text{ vol.} & & 2\text{ vol.} \\ \end{matrix}$

(i) Given,

1 vol. of di nitrogen produced at the same time as 8.96 litres of steam (2 vol)

Hence, Vol of di nitrogen = $\dfrac{8.96}{2}$ = 4.48 lit.

Hence, volume of di nitrogen oxide produced = 4.48 lit.

(ii) 2 vol = (2 x 22.4) lit steam is produced from 80 g NH4NO3

∴ 8.96 lit of steam will be produced by $\dfrac{80}{2 \times 22.4}$ x 8.96 = 16 g

Hence, 16 g of ammonium nitrate is required to be heated.

(iii) % of oxygen in ammonium nitrate = $\dfrac{3 \times 16}{80}$ x 100 = 60 %

#### Question 44

Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide? The equation for the reaction is:

3CuO + 2NH3 ⟶ 3Cu + 3H2O + N2

$\begin{matrix} 3\text{CuO} & + & 2\text{NH}_3 \\ 3\text{ mol.} & & 2 \text{ mol.} \\ = 3 \times 80 \text{ g.} & & = 2 \times 22.4 \text{ lit.} \\ = 240 \text{ g.} & & = 44.8\text{ lit.} \\ \end{matrix}$

$\longrightarrow 3\text{Cu} + 3\text{H}_2\text{O} + \text{N}_2$

240 g of CuO is reduced by 44.8 lit of NH3

∴ 120 g of CuO is reduced by $\dfrac{44.8}{240}$ x 120 = 22.4 lit.

Hence, 22.4 lit of NH3 is required

#### Question 45

(a) Calculate the number of moles and the number of molecules present in 1.4 g of ethylene gas. What is the volume occupied by the same amount of ethylene?

(b) What is the vapour density of ethylene?

Gram molecular mass of C2H4
= 2(12) + 4(1)
= 24 + 4 = 28 g

As,

28 g of C2H4 = 1 mole
∴ 1.4 g of C2H4 = $\dfrac{1}{28}$ x 1.4 = 0.05 moles

1 mole = 6 × 1023 molecules
∴ 0.05 moles = 6 × 1023 x 0.05 = 3 x 1022 molecules

Hence, no. of moles is 0.05 and no. of molecules is 3 x 1022.

Vol. occupied by 1 mole = 22.4 lit
∴ Vol. occupied by 0.05 moles = 22.4 x 0.05 = 1.12 lit.

Hence, vol. occupied is 1.12 lit

(b)
$\text{Vapour density} = \dfrac{\text{Molecular weight}}{2} = \dfrac{28}{2} = 14$

Hence, vapour density is 14

#### Question 46

(a) Calculate the percentage of sodium in sodium aluminium fluoride (Na3AlF6) correct to the nearest whole number.

(F = 19; Na =23; Al = 27)

(b) 560 ml of carbon monoxide is mixed with 500 ml of oxygen and ignited. The chemical equation for the reaction is as follows:

2CO + O2 ⟶ 2CO2

Calculate the volume of oxygen used and carbon dioxide formed in the above reaction.

Molecular weight of sodium aluminium fluoride (Na3AlF6)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 69 g of Na

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{69}{210}$ x 100 = 32.85% = 33%

Hence, percentage of sodium in sodium aluminium fluoride (Na3AlF6) is 33%

(b)

$\begin{matrix} 2\text{CO} & + & \text{O}_2 & \longrightarrow & 2\text{CO}_2 & \\ 2 \text{ vol.} & : & 1 \text{ vol.} & \longrightarrow & 2\text{ vol.} \\ \end{matrix}$

1 mole of O2 has volume = 22400 ml

Volume of oxygen used by 2 × 22400 ml CO = 22400 ml

∴ Vol. of O2 used by 560 ml CO

= $\dfrac{22400}{2 × 22400}$ x 560

= 280 ml

Volume of CO2 formed by 2 × 22400 ml CO = 2 x 22400 ml

∴ Vol. of CO2 formed by by 560 ml CO

= 560 ml

#### Question 47a(2009)

A gas cylinder of capacity of 20 dm3 is filled with gas X the mass of which is 10 g. When the same cylinder is filled with hydrogen gas at the same temperature and pressure the mass of the hydrogen is 2 g, hence the relative molecular mass of the gas is:

i. 5

ii. 10

iii. 15

iv. 20

10

Working

Vapour density of gas =

$\dfrac{\text{Wt. of certain volume of gas }}{\text {Wt. of same volume of H}_2} \\[0.5em] = \dfrac{10}{2} \\[0.5em] = 5 \text{ g}$

Molecular weight = 2 x Vapour density
= 2 x 5 = 10 g

Hence, relative molecular mass of gas is 10 g

#### Question 47(b-i)(2009)

Calcium carbide is used for the artificial ripening of fruits. Actually the fruit ripens because of the heat evolved while calcium carbide reacts with moisture. During this reaction calcium hydroxide and acetylene gas is formed. If 200 cm3 of acetylene is formed from a certain mass of calcium carbide, find the volume of oxygen required and carbon dioxide formed during the complete combustion. The combustion reaction can be represented as below.

2C2H2(g) + 5O2(g) ⟶4CO2(g) + 2H2O(g)

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 \longrightarrow & 4\text{CO}_2 & + &2\text{H}_2\text{O} \\ 2\text{ mol.} & & 5 \text{ mol.} & 4\text{ mol.}\ \end {matrix}$ According to Gay-Lussac's law,

2 volume of acetylene requires 5 volume of oxygen

1 volume of acetylene requires $\dfrac{5}{2}$ = 2.5 volume of oxygen

∴ 200 cm3 of acetylene requires $\dfrac{2.5}{1}$ x 200

= 500 cm3 of oxygen

2 volume of acetylene produces 4 volume of CO2

1 volume of acetylene produces $\dfrac{4}{2}$ = 2 volume of CO2

∴ 200 cm3 produces $\dfrac{2}{1}$ x 200 = 400 cm3 of CO2

#### Question 47(b-ii)(2009)

A gaseous compound of nitrogen and hydrogen contains 12.5% hydrogen by mass. Find the molecular formula of the compound if its relative molecular mass is 37. (N = 14, H = 1).

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
Nitrogen100 - 12.5 = 87.514$\dfrac{87.5}{14}$ = 6.25$\dfrac{6.25}{6.25}$ = 1
Hydrogen12.51$\dfrac{12.5}{1}$ = 12.5$\dfrac{12.5}{6.25}$ = 2

Simplest ratio of whole numbers N : H = 1 : 2

Hence, empirical formula is NH2

Empirical formula weight = 14 + 2(1) = 16

Relative molecular mass = 37

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{37}{16} = 2.31 = 2$

∴ Molecular formula = n[E.F.] = 2[NH2] = N2H4

#### Question 47(c-i)(2009)

A gas cylinder contains 24 × 1024 molecules of nitrogen gas. If Avogadro's number is 6 × 1023 and the relative atomic mass of nitrogen is 14, calculate :

(1) Mass of nitrogen gas in the cylinder.

(2) Volume of nitrogen at STP in dm3

(1) Gram molecular mass of N2 = 2(14) = 28 g

6 × 1023 molecules of N2 weighs 28 g

∴ 24 × 1024 molecules will weigh $\dfrac{28}{6 \times 10^{23}}$ x 24 × 1024 = 1120 g

(2) 6 × 1023 molecules of N2 occupies 22.4 dm3

∴ 24 × 1024 molecules of N2 will occupy $\dfrac{22.4}{6 \times 10^{23}}$ x 24 × 1024 = 896 dm3

#### Question 47(c-ii)(2009)

Commercial sodium hydroxide weighing 30g has some sodium chloride in it. The Mixture on dissolving in water and subsequent treatment with excess silver nitrate solution formed a precipitate weighing 14.3 g. What is the percentage of sodium chloride in the commercial sample of sodium hydroxide? The equation for the reaction is

NaCl + AgNO3 ⟶ AgCl + NaNO3

(Relative molecular mass of NaCl = 58; AgCl = 143)

$\underset{58 \text{g}}{\text{NaCl}} + \text{AgNO}_3 \longrightarrow \underset{143 \text{ g}}{\text{AgCl}} + \text{NaNO}_3$

(i) 143 g AgCl is formed by 58 g NaCl

∴ 14.3 g of AgCl will be formed by $\dfrac{58}{143}$ x 14.3 = 5.8 g

Percentage of NaCl = $\dfrac{5.8}{30}$ x 100 = 19.33%

Hence, percentage of NaOH is 19.33%

#### Question 47(c-iii)(2009)

A certain gas 'X' occupies a volume of 100 cm3 at S.T.P. and weighs 0.5 g. Find its relative molecular mass

100 cm3 weighs 0.5 g

∴ 22400 cm3 weighs $\dfrac{0.5}{100}$ x 22400 = 112 g

Hence, relative molecular mass of the gas = 112 g.

#### Question 48(a-i)(2010)

LPG stands for liquefied petroleum gas. Varieties of LPG are marketed including a mixture of propane (60%) and butane (40%). If 10 litre of this mixture is burnt, find the total volume of carbon dioxide gas added to the atmosphere. Combustion reactions can be represented as:

C3H8 + 5O2 ⟶ 3CO2 + 4H2O

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

Given, 10 litres of this mixture contains 60% propane and 40% butane. Hence, propane is 6 litres and butane is 4 litres

$\begin{matrix} \text{C}_3\text{H}_8 & + & 5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 1 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 3\text{ vol.} \\ 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} \\ \end{matrix}$

1 Vol. C3H8 produces carbon dioxide = 3 Vol

So, 6 litres C3H8 will produce carbon dioxide = 3 x 6 = 18 litres

2 Vol. C4H10 produces carbon dioxide = 8 Vol

So, 4 litres C4H10 will produce carbon dioxide = $\dfrac{8}{2}$ x 4 = 16 litres

Hence, 34 (i.e., 18 + 16) litres of CO2 is produced.

#### Question 48(a-ii)(2010)

Calculate the percentage of nitrogen and oxygen in ammonium nitrate. (Relative molecular mass of ammonium nitrate is 80, H = 1, N = 14, O = 16).

Given,

molecular mass of ammonium nitrate = 80

mass of nitrogen in 1 mole of ammonium nitrate (NH4NO3) is 2(14) = 28 g

percentage of nitrogen in ammonium nitrate = $\dfrac{28}{80}$ x 100 = 35%

mass of oxygen in 1 mole of ammonium nitrate (NH4NO3) is 3(16) = 48 g

percentage of oxygen in ammonium nitrate = $\dfrac{48}{80}$ x 100 = 60%

Hence, percentage of nitrogen is 35% and percentage of oxygen is 60%

#### Question 48b(2010)

4.5 moles of calcium carbonate are reacted with dilute hydrochloric acid.

(i) Write the equation for the reaction.

(ii) What is the mass of 4.5 moles of calcium carbonate? (Relative molecular mass of calcium carbonate is 100)

(iii) What is the volume of carbon dioxide liberated at STP?

(iv) What mass of calcium chloride is formed? (Relative molecular mass of calcium chloride is 111).

(v) How many moles of HCl are used in this reaction?

(i) CaCO3 + 2HCl ⟶ CaCl2 + H2O + CO2

(ii) Given,

1 mole of CaCO3 = molecular mass of CaCO3 = 100 g

∴ 4.5 moles of CaCO3 weighs $\dfrac{100}{1}$ x 4.5 = 450 g

(iii) 1 mole of CaCO3 produces 1 mole of CO2 and 1 mole occupies 22.4 l of volume.

∴ 4.5 moles of CaCO3 will produce 4.5 moles of CO2 and 4.5 moles will occupy 22.4 x 4.5 = 100.8 L

(iv) 1 mole CaCO3 produces 111 g. CaCl2

∴ 4.5 moles of CaCO3 will produce = 111 x 4.5 = 499.5 g.

(v)

$\begin{matrix}\text{CaCO}_3 & : & \text{HCl} & \\ \text{ 1 mole} & : & 2 \text{ moles} \\ \text{ 4.5 mole} & : & x \text{ moles} \\ \end{matrix}$

∴ number of moles of HCl used = 2 x 4.5 = 9 moles.

Hence, moles of HCl used = 9 moles.

#### Question 49a(2011)

i. Calculate the volume of 320 g of SO2 at STP. (Atomic mass: S = 32 and O = 16).

ii. State Gay-Lussac's Law of combining volumes

iii. Calculate the volume of oxygen required for the complete combustion of 8.8 g of propane (C3H8). (Atomic mass: C = 12, O = 16, H = I, Molar Volume = 22.4 dm3 at stp.)

(i) Gram molecular mass of SO2 = 32 + 2(16) = 32 + 32 = 64 g

64 g of SO2 occupy 22.4 lit of vol.

320 g of SO2 will occupy = $\dfrac{22.4}{64}$ x 320 = 112 lit.

Hence, volume of 320 g of SO2 = 112 lit.

(ii) Gay-Lussac's law Gay-Lussac's Law states "When gases react, they do so in volumes which bear a simple ratio to one another and to the volume of the gaseous product, if all the volumes are measured at the same temperature and pressure."

(iii)

$\begin{matrix} \text{C}_3\text{H}_8 & + &5\text{O}_2 & \longrightarrow & 3\text{CO}_2 & + & 4\text{H}_2\text{O} \\ 3(12) + 8(1) & & 5 \text{ vol} \\ = 44 \text{ g}& & 5 (22.4) \text{ lit} \\ \end{matrix}$

(i) 44 g propane requires 5 x 22.4 lit of oxygen

∴ 8.8 g of propane will require $\dfrac{5 \times 22.4}{44}$ x 8.8 = 22.4 lit.

Hence, 22.4 lit of Oxygen is required.

#### Question 49(b-i)(2011)

An organic compound with vapour density = 94 contains C = 12.67%, H = 2.13%, and Br = 85.11%. Find the molecular formula. (Atomic mass : C = 12, H = 1, Br = 80)

Element% compositionAt. wt.Relative no. of atomsSimplest ratio
C12.6712$\dfrac{12.67}{12}$ = 1.05$\dfrac{1.05}{1.05}$ = 1
H2.131$\dfrac{2.13}{1}$ = 2.13$\dfrac{2.13}{1.05}$ = 2.02 = 2
Br85.1180$\dfrac{85.11}{80}$ = 1.06$\dfrac{1.06}{1.05}$ = 1

Simplest ratio of whole numbers = C : H : Br = 1 : 2 : 1

Hence, empirical formula is CH2Br

Empirical formula weight = 12 + 2(1) + 80 = 94

Vapour density (V.D.) = 94

Molecular weight = 2 x V.D. = 2 x 94

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 94}{94} = 2$

∴ Molecular formula = n[E.F.] = 2[CH2Br] = C2H4Br2

#### Question 49(b-ii)(2011)

Calculate the mass of:

1. 1022 atoms of sulphur.

2. 0.1 mole of carbon dioxide.

(i) gram molecular mass of S = 32

6 × 1023 atoms weigh = 32 g

1022 atoms will weigh = $\dfrac{32}{6 \times 10^{23}}$ x 1022 = 0.533 g

(ii) 1 mole of carbon dioxide weighs = C + 2(O) = 12 + 2(16) = 12 + 32 = 44 g

∴ 0.1 mole of carbon dioxide weighs = 44 x 0.1 = 4.4 g.

#### Question 50a(2012)

Concentrated nitric acid oxidises phosphorus to phosphoric acid according to the following equation:

P + 5HNO3 [conc.] ⟶ H3PO4 + H2O + 5NO2.

If 9.3 g of phosphorus was used in the reaction, calculate :

(i) Number of moles of phosphorus taken.

(ii) The mass of phosphoric acid formed.

(iii) The volume of nitrogen dioxide produced at STP.

$\begin{matrix} \text{P} & + & 5\text{HNO}_3 & \longrightarrow & \text{H}_3\text{PO}_4 & + & \text{H}_2\text{O} & + & 5\text{NO}_2 \\ 31 \text{ g} & & [conc.] & & 3(1) + 31 \\ & & & & + 4(16) \\ & & & & = 98 \text{ g} \\ \end{matrix}$

(i)

31 g of P = 1 mole

∴ 9.3 g of P = $\dfrac{1}{31}$ x 9.3 = 0.3 moles.

Hence, 0.3 moles of phosphorous was taken for the reaction.

(ii)

31 g of P forms 98 g of phosphoric acid

∴ 9.3 g will form $\dfrac{ 98}{31}$ x 9.3 = 29.4 g.

Hence, 29.4 g. of phosphoric acid is formed

(iii)

31 g of P produces 5 vol = 5 x 22.4 lit.

∴ 9.3 g will produce = $\dfrac{5 \times 22.4}{31}$ x 9.3 = 33.6 lit.

#### Question 50(b-i)(2012)

67.2 litres of hydrogen combines with 44.8 litres of nitrogen to form ammonia :

N2(g) + 3H2(g) ⟶ 2NH3(g).

Calculate the vol. of ammonia produced. What is the substance, if any, that remains in the resultant mixture ?

[By Lussac's law]

$\begin{matrix} \text{N}_2 & + & 3\text{H}_2 & \longrightarrow & 2 \text{NH}_3 \\ 1 \text{ vol.} & : & 3 \text{ vol.} & \longrightarrow & 2\text{ vol.} \end{matrix}$

To calculate the volume of ammonia gas formed.

$\begin{matrix} \text{H}_2 & : & \text{NH}_3 \\ 3 & : & 2 \\ 67.2 & : & x \end{matrix}$

$\therefore \dfrac{2}{3} \times 67.2 = 44.8 \text{ lit}$

Hence, volume of NH3 formed is 44.8 lit.

We know,

$\begin{matrix}\text{H}_2 & : & \text{N}_2 \\ 3 & : & 1 \\ 67.2 & : & x \end{matrix}$

$\therefore \dfrac{1}{3} \times 67.2 = 22.4 \text{ lit}$

∴ nitrogen left = 44.8 - 22.4 = 22.4 lit.

Hence, 22.4 lit of nitrogen remains in the resultant mixture.

#### Question 50(b-ii)(2012)

The mass of 5.6 dm3 of a certain gas at STP is 12.0 g. Calculate the relative molecular mass of the gas.

5.6 dm3 weighs 12 g

∴ 22.4 dm3 weighs $\dfrac{12}{5.6}$ x 22.4 = 48 g

Hence, relative molecular mass of the gas = 48 amu

#### Question 50(b-iii)(2012)

Find the total percentage of Magnesium in magnesium nitrate crystals, Mg(NO3)2.6H2O.

(Mg = 24, N = 14, 0 = 16 and H = 1)

Molecular weight of Mg(NO3)2.6H2O = 24 + 2[14 + 3(16)] + 6[2(1) + 16]
= 24 + 2[14 + 48] + 6[18]
= 24 + 2(62) + 108
= 24 + 124 + 108 = 256 g

256 g of magnesium nitrate crystals contains 24 g of magnesium

∴ 100 g of magnesium nitrate crystals will contain = $\dfrac{24}{256}$ x 100 = 9.375% = 9.38%

Hence, percentage of magnesium in magnesium nitrate crystals is 9.38%

#### Question 51(a-i)(2013)

What volume of oxygen is required to burn completely 90 dm3 of butane under similar conditions of temperature and pressure?

2C4H10 + 13O2 ⟶ 8CO2 + 10H2O

[By Lussac's law]

$\begin{matrix} 2\text{C}_4\text{H}_{10} & + & 13\text{O}_2 & \longrightarrow & 8\text{CO}_2 & + & 10\text{H}_2\text{O} \\ 2 \text{ vol.} & : & 13 \text{ vol.} & \longrightarrow & 8\text{ vol.} & : & 10\text{ vol.} \end{matrix}$

To calculate the volume of oxygen

$\begin{matrix}\text{C}_4\text{H}_{10} & : & \text{O}_2 \\ 2 & : & 13 \\ 90 & : & x \end{matrix}$

$\dfrac{13}{2} \times 90 = 585 \text{ dm}^3$

Hence, vol of oxygen = 585 dm3

#### Question 51(a-ii)(2013)

The vapour density of a gas is 8. What would be the volume occupied by 24.0 g of the gas at STP?

Given, V.D. = 8

Gram molecular mass = V.D. x 2 = 8 x 2 = 16 g.

16 g occupies 22.4 lit.

∴ 24 g. will occupy = $\dfrac{22.4}{16}$ x 24 = 33.6 lit.

Hence, volume occupied by gas = 33.6 lit.

#### Question 51(a-iii)(2013)

A vessel contains X number of molecules of hydrogen gas at a certain temperature and pressure. How many molecules of nitrogen gas would be present in the same vessel under the same conditions of temperature and pressure?

According to Avogadro's law, equal volume of all gases under similar conditions of temperature and pressure contain equal number of molecules.

Hence, number of molecules of N2 = Number of molecules of H2 = X

#### Question 51(b)(2013)

O2 is evolved by heating KClO3 using MnO2 as a catalyst.

2KClO3 $\xrightarrow{\text{MnO}_2}$ 2KCl + 3O2

(i) Calculate the mass of KClO3 required to produce 6.72 litre of O2 at STP.

(atomic masses of K = 39, Cl = 35.5, 0 = 16).

(ii) Calculate the number of moles of oxygen present in the above volume and also the number of molecules.

(iii) Calculate the volume occupied by 0.01 mole of CO2 at STP.

$\begin{matrix} 2\text{KClO}_3 & \xrightarrow{MnO_2} & 2\text{KCl} & + & 3\text{O}_2 \\ 2[39 + 35.5 & & 2[39 & & 3(22.4) \\ + 3(16)] & & + 35.5] & & = 67.2 \text{ lit.} \\ = 245 \text{ g} & & = 149 \text{g} \\ \end{matrix}$

(i)

67.2 lit. of O2 is produced by 245 g of KClO3

∴ 6.72 lit of O2 will be obtained from $\dfrac{245}{67.2}$ x 6.72 = 24.5 g

(ii)

22.4 lit = 1 mole
∴ 6.72 lit = $\dfrac{1}{22.4}$ x 6.72 = 0.3 moles

1 mole = 6.023 x 1023 molecules
∴ 0.3 moles = 0.3 x 6.023 x 1023 = 1.806 x 1023 molecules.

(iii)

Volume occupied by 1 mole of CO2 at STP = 22.4 litres

So, volume occupied by 0.01 mole of CO2 at STP
= 22.4 × 0.01
= 0.224 litres

#### Question 52(a-i)(2014)

Oxygen oxidises ethyne to carbon dioxide and water as shown by the equation:

2CO2H2 + 5O2 ⟶ 4CO2 + 2H2O

What volume of ethyne gas at s.t.p. is required to produce 8.4 dm3 of carbon dioxide at STP ? [H = 1, C = 12, 0 = 16]

[By Lussac's law]

$\begin{matrix} 2\text{C}_2\text{H}_2 & + & 5\text{O}_2 &\longrightarrow & 4\text{CO}_2 & + & 2\text{H}_2\text{O} & \\ 2 \text{ vol.} & : & 5 \text{ vol.} & \longrightarrow & 4\text{ vol.} & : & 2\text{ vol.} \end{matrix}$

To calculate the volume of ethyne gas

$\begin{matrix}\text{CO}_2 & : & \text{C}_2\text{H}_2 \\ 4 & : & 2 \\ 8.4 & : & x \end{matrix}$

$\dfrac{2}{4} \times 8.4 = 4.2 \text{ dm}^3$

Hence, volume of ethyne gas required = 4.2 dm3.

#### Question 52(a-ii)(2014)

A compound made up of two elements X and Y has an empirical formula X2 Y. If the atomic weight of X is 10 and that of Y is 5 and the compound has a vapour density (V.D.) 25, find it's molecular formula.

Empirical formula is X2Y

Empirical formula weight = 2(10) + 5 = 25

Vapour density (V.D.) = 25

Molecular weight = 2 x V.D. = 2 x 25

$\text{n} = \dfrac{\text{Molecular weight}}{\text{Empirical formula weight}} \\[0.5em] = \dfrac{2 \times 25}{25} = 2$

∴ Molecular formula = n[E.F.] = 2[X2Y] = X4Y2

#### Question 52(b) (2014)

A cylinder contains 68 g of Ammonia gas at STP

(i) What is the volume occupied by this gas?

(ii) How many moles of ammonia are present in the cylinder?

(iii) How many molecules of ammonia are present in the cylinder?

By Avogadro's law: Under the same conditions of temperature and pressure equal volumes of all gases contain the same number of molecules.

(i) Gram molecular mass of ammonia = N + 3(H) = 14 + 3 = 17 g.

17 g. occupies 22.4 lit. of vol.

∴ 68 g will occupy = $\dfrac{22.4}{17}$ x 68 = 89.6 lit.

Hence, volume occupied by this gas = 89.6 lit.

(ii) 17 g = 1 mole

∴ 68 g = $\dfrac{1}{17}$ x 68 = 4 moles.

1 mole = 6 × 1023

∴ 4 moles = 4 x 6.023 × 1023 molecules.

Hence, Moles = 4

(iii) molecules = 4 x 6.023 × 1023 = 2.4 x 1024 molecules