The Language of Chemistry

Questions

Question 1(1985)

XCl₂ is the chloride of a metal X. State the formula of the sulphate and the hydroxide of the metal X.

Answer

Chloride of a metal X is XCl₂

By interchanging subscript and writing as superscript:

$\underset{\phantom{1}{1}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{Cl}} \Rightarrow \overset{\phantom{2}{2}}{\text{X}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{Cl}} \\[0.5em]$

Therefore, valency of metal X = 2.

Formula of the sulphate:

$\text{X}^{2+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{S}}\text{O}_{4} \\[0.5em]$

As valency of both X and SO₄ is 2 so dividing by 2 we get 1 but 1 is never written so we get the formula as $\bold{XSO}_{\bold{4}}$

Formula of the hydroxide:

$\text{X}^{2+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{2}{\text{O}}\text{H} \\[0.5em]$

Dropping 1 and enclosing OH in brackets, we get the formula as $\bold{X(OH)}_{\bold{2}}$

Therefore, we get

Formula of Sulphate : $\bold{XSO}_{\bold{4}}$

Formula of Hydroxide : $\bold{X(OH)}_{\bold{2}}$

Question 1(1987)

An element X is trivalent. Write the balanced equation for the combustion of X in oxygen.

Answer

Valency of X is 3+ and that of Oxygen is 2-

By interchanging the valency number and shifting it to the lower right side of the atom or radical, we get the formula : X₂O₃

Equation for the combustion of X in oxygen: 4X + 3O₂ ⟶ 2X₂O₃

Question 1(1991)

The formula of the nitride of a metal X is XN, state the formula of :

(i) it's sulphate (ii) it's hydroxide.

Answer

Nitride of a metal X is XN.

Since valency of nitrogen is 3-
so valency of X is 3+

Formula of the sulphate:

$\text{X}^{3+} \phantom{\nearrow} \text{SO}_{4}^{2-} \\[0.5em] \overset{\phantom{3}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{2}{\text{S}}\text{O}_{4} \Rightarrow \underset{\phantom{2}{2}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{S}}\text{O}_{4} \\[0.5em]$

So, we get the formula as $\bold{X}_2\bold{(SO}_{\bold{4}}\bold{)}_\bold{3}$

Formula of the hydroxide:

$\text{X}^{3+} \phantom{\nearrow} \text{OH}^{1-} \\[0.5em] \overset{\phantom{3}{3}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \overset{1}{\text{O}}\text{H} \Rightarrow \underset{\phantom{1}{1}}{\text{X}} \space {\searrow}\mathllap{\swarrow} \space \underset{3}{\text{O}}\text{H} \\[0.5em]$

Dropping 1, we get the formula as $\bold{X(OH)}_{\bold{3}}$

Therefore, we get

Formula of Sulphate : $\bold{X}_2\bold{(SO}_{\bold{4}}\bold{)}_\bold{3}$

Formula of Hydroxide : $\bold{X(OH)}_{\bold{3}}$

Question 1(1992)

What is the valency of nitrogen in :

(i) NO

(ii) N₂O

(iii) NO₂

Answer

$\text{(i)} \space \text{NO} = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]$

But valency of O is 2. Multiplying by 2, we get:

$\overset{{2 \times 1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]$

Therefore, valency of Nitrogen is 2.

$\text{(ii)} \space \text{N}_2\text{O} = \underset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{1}{\text{O}} \Rightarrow \overset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]$

Therefore, valency of Nitrogen is 1.

$\text{(ii)} \space \text{NO}_2 = \underset{\phantom{1}{1}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]$

But valency of O is 2. Multiplying by 2, we get:

$\overset{{2 \times 2}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{N}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]$

Therefore, valency of Nitrogen is 4.

Additional Questions

Question 1

What is meant by the term 'symbol'. Give the qualitative and quantitative meaning of the term 'symbol'.

Answer

Symbol represents the short form of an element.

Qualitative meaning — A symbol represents specific element or one atom of an element. E.g., 'S' represents one atom of the element sulphur.

Quantitative meaning — A symbol also represents the weight of the element equal to it's atomic weight i.e. it represents how many times an atom is heavier than one atomic mass unit [a.m.u.] which is defined as $\dfrac{1}{12}$th the mass of a carbon atom C¹²

Question 2

Name three metals whose symbols are derived from :

(a) the first letter of the name of the element

(b) from their Latin names.

Answer

(a) Elements whose symbols are derived from first letter of the name of the element :

• Carbon - C
• Sulphur - S
• Oxygen - O

(b) Elements whose symbols are derived from their Latin names :

• Kalium [Potassium] - K
• Ferrum [Iron] - Fe
• Natrium [Sodium] - Na

Question 3

Explain the meaning of the term 'valency'. State why the valency of the metal potassium is +1 and of the non-metal chlorine is -1.

Answer

Valency is the number of hydrogen atoms which can combine with or displace one atom of the element or radical so as to form a compound.

Valency of a metal is the number of electrons lost per atom of the metal. Valency of all metals and hydrogen is considered positive. Therefore, valency of potassium is 1+ as it has 1 electron in outer most shell which it loses and becomes K⁺.

Valency of non-metal is the number of electrons gained per atom of the non-metal. Valency of all non-metals/groups of non-metals is taken as negative. Therefore, valency of non-metal chlorine is -1, because chlorine [Cl] has 7 electrons in valence shell and gains 1 electron and becomes [Cl^–].

Question 4

What is meant by the term 'variable valency'. Give a reason why silver exhibits a valency of +1 and +2.

Answer

Certain elements exhibit more than one valency hence it is said that these elements have variable valency.

Reasons for exhibiting variable valency — An atom of an element can sometimes lose more electrons than are present in it's valence shell. This happens when it loses electrons from the penultimate [i.e., last but one] shell and hence exhibit more than one or variable valency.

Variable valency of Silver — Atomic number of Silver (Ag) is 47. Its electronic configuration is [2, 8, 18, 18, 1]. The outermost shell has one electron and the penultimate shell has 18 electrons. However, the penultimate shell has not attained stability and one more electron sometimes jumps to the outermost shell there by increasing valence electrons and the new configuration [2, 8, 18, 17, 2] loses two electrons and has valency [+2]. Therefore, Silver exhibits varibale valency forming Ag¹⁺ and Ag²⁺.

Question 5

Give examples of eight metals which show variable valency. State the valency of sulphur in :

(a) SO₂
(b) SO₃

Answer

Eight metals which show variable valency are :

(a) Valency of sulphur in SO₂ is 4

By interchanging subscript and writing as superscript:

$\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{2}{\text{O}} \Rightarrow \overset{\phantom{2}{2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]$

But valency of O is 2. Multiplying by 2, we get:

$\overset{{2 \times 2}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{4}{4}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]$

Therefore, valency of Sulphur is 4.

(b) Valency of sulphur in SO₃ is 6

$\underset{\phantom{1}{1}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \underset{3}{\text{O}} \Rightarrow \overset{\phantom{3}{3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{1}{\text{O}} \\[0.5em]$

But valency of O is 2. Multiplying by 2, we get:

$\overset{{2 \times 3}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2 \times 1}{\text{O}} \Rightarrow \overset{\phantom{6}{6}}{\text{S}} \space {\nearrow}\mathllap{\nwarrow} \space \overset{2}{\text{O}} \\[0.5em]$

Therefore, valency of Sulphur is 6.

Question 6

State the valency in each case and name the following elements or radicals given below :

1. K
2. Cr₂O₇
3. Cl
4. Ni
5. ClO₃
6. CO₃
7. Ba
8. HCO₃
9. NO₂
10. Na
11. Br
12. Zn
13. Mg
14. O
15. Co
16. CrO₄
17. ClO
18. MnO₄
19. Li
20. I
21. OH
22. O₂
23. ZnO₂
24. SiO₃
25. NO₃
26. SO₃
27. SO₄
28. PO₄
29. N
30. C
31. PO₃
32. Al
33. Ca
34. H
35. PbO₂
36. HSO₃
37. AlO₂
38. Cr
39. HSO₄
40. NH₄

Answer

Question 7

State the variable valencies of the following elements and give their names. (a) Cu, (b) Ag, (c) Hg, (d) Fe, (e) Pb, (f) Sn, (g) Mn, (h) Pt, (i) Au

Answer

Question 8

State which of the following elements or radicals are divalent –

(a) Lithium, (b) Nickel, (c) Ammonium, (d) Bromide, (e) Sulphite, (f) Nitride, (g) Carbide, (h) Chromium, (i) Bisulphite, (j) Dichromate, (k) Permanganate.

Answer

The divalent elements or radicals are:

• Nickel
• Sulphite
• Dichromate

Question 9

Explain the meaning of the term 'compound' with a suitable example. State the main characteristics of a compound with special reference to the compound iron [II] sulphide.

Answer

Compound is a pure substance composed of two or more elements combined chemically in a fixed proportion. For example, water (H₂O) is composed of two elements — Hydrogen and Oxygen in a fixed proportion by weight.

Main characteristics - with special reference to the compound iron [II] sulphide:

1. Components are present in - a definite proportion.

• Elements iron and sulphur combine in a definite proportion.

2. Compound is always homogenous.
3. Particles in a compound are of one kind.

• Composition of iron [II] sulphide, - cannot be varied, hence is uniform composition.
• Component - may not be seen separately.

4. Compound [iron [II] sulphide] has a definite set of properties.
5. Components in [Iron [II] Sulphide] do not retain their original properties and can be separated only by chemical means.

• Particles in iron [II] sulphide are chemically combined and hence -
• iron in iron [II] sulphide cannot be attracted by a magnet and does not give H₂ with dil. acid.
• sulphur present in iron [II] sulphide - is insoluble in solvent CS₂.

Question 10

Name the elements in the compound and give the formula – of the following compounds :

(a) Nitric acid, (b) Carbonic acid, (c) Phosphoric acid, (d) Acetic acid, (e) Blue vitriol, (f) Green vitriol, (g) Glauber's salt, (h) Ethane, (i) Ethanol

Answer

Question 11

Explain the term 'chemical formula'. State why the molecular formula of zinc carbonate is ZnCO₃

Answer

A molecule of a substance i.e., element or compound could be represented by symbols. This representation is known as chemical formula.

As molecular formula also indicates the number of units of the radicals present in a compound with the proper subscript outside the unit of the radical and in the case of ZnCO₃, there are 1 atom of Zn and C whereas 3 atoms of O. Therefore, we get ZnCO₃.

Question 12(1)

Write the formula of the following compounds :

Potassium

Answer

Potassium

Question 12(2)

Write the formula of the following compounds :

Sodium

Answer

Sodium

Question 12(3)

Write the formula of the following compounds :

Calcium

Answer

Question 12(4)

Write the formula of the following compounds :

Magnesium

Answer

Question 12(5)

Write the formula of the following compounds :

Zinc

Answer

Question 12(6)

Write the formula of the following compounds :

Aluminium

Answer

Question 12(7)

Write the formula of the following compounds :

Copper

Answer

Question 12(8)

Write the formula of the following compounds :

Iron

Answer

Question 12(9)

Write the formula of the following compounds :

Lead

Answer

Question 12(10)

Write the formula of the following compounds :

Silver

Answer

Question 13

Write the names of the following compounds :

Answer

Question 14

Explain the term 'chemical equation'. What is meant by 'reactants' and 'products' in a chemical equation.

Answer

Chemical equation is a shorthand form for a chemical change. It shows the result of a chemical change in which the reactants and the products are represented by:

• symbols [in the case of elements]
• formula [in the case of compounds]

The substances which take part in a chemical change and are written on the left side of the equation are called the reactants.

The substances formed as a result of a chemical change and are written on the right side of the equation are called the products.

E.g., NH₃ + H₂O ⟶ NH₄OH

Here, NH₃ and H₂O are reactants and NH₄OH is the product.

Question 15

Give an example of a chemical equation in which two reactants form –

(a) one product
(b) two products
(c) three products
(d) four products

Answer

(a) NH₃ + H₂O ⟶ NH₄OH

(b) C₂H₅Br + NaOH ⟶ C₂H₅OH + NaBr

(c) 6CO₂ + 12H₂O ⟶ C₆H₁₂O₆ + 6H₂O + 6O₂

(d) 2KMnO₄ + 16HCl ⟶ 2KCl + 2MNCl₂ + 8H₂O + 5Cl₂

Question 16

$2\text{KClO}_3 \xrightarrow{\text{MnO}_2} 2\text{KCl} + 3\text{O}_2[\text{g}]$ is a balanced equation.

(a) State what is a 'balanced equation'.

(b) Give a reason why the above equation is balanced.

(c) State why the compound MnO₂ is written above the arrow.

Answer

(a) A balanced equation is one in which the number of atoms of each element is the same on the side of the reactants and on the side of the products.

(b) As atoms of each reactant [K, Cl, O] on the L.H.S. is equal to the number of the atoms of products on the R.H.S., hence the given equation is balanced.

(c) MnO₂ is a catalyst [simply increases the rate of reaction] and it does not take part in the reaction and so it is written on the top of the arrow.

Question 17

What do the following symbols present in a chemical equation mean.

(i) ⟶

(ii) ⇌

(iii) (s)

(iv) (I)

(v) (g)

(vi) (aq.)

Answer

(i) ⟶ : The direction of the reaction is irreversible

(ii) ⇌ : The direction of the reaction is reversible

(iii) (s) : The state of the matter is solid

(iv) (I) : The state of the matter is liquid

(v) (g) : The state of the matter is gas

(vi) (aq.) : The substance is in aqueous form.

Question 18

CaCO₃ + 2HCl [dil.] ⟶ CaCl₂ + H₂O + CO₂ [g]

(a) State the information provided by the above chemical equation.

(b) State the information not conveyed by the above chemical equation.

Answer

(a) Information provided by the equation:

• One molecule of calcium carbonate reacts with two molecules of [dil.] hydrochloric acid to produce one molecule of calcium chloride, one molecule of water and one molecule of carbon dioxide.
• The reaction is irreversible.
• Hydrochloric acid is in dil. form.
• Carbon dioxide produced is in gaseous form.

(a) Information not provided by the equation:

• The concentration of the reactants and products.
• The speed of the reaction
• Change in colour occurring
• Completion of the reaction i.e., whether the reaction is complete or not
• Changes in evolution of light or sound energy during the occurrence of the reaction.

Question 19

Balance the following simple equations :

1. C + O₂ ⟶ CO

2. N₂ + O₂ ⇌ NO

3. ZnS + O₂ ⟶ ZnO + SO₂

4. Al + O₂ ⟶ Al₂O₃

5. Mg + N₂ ⟶ Mg₃N₂

6. Al + N₂ ⟶ AlN

7. NO + O₂ ⟶ NO₂

8. SO₂ + O₂ ⇌ SO₃

9. H₂ + Cl₂ ⟶ HCl

10. Fe + Cl₂ ⟶ FeCl₃

11. H₂S + Cl₂ ⟶ S + HCl

12. FeCl₂ + Cl₂ ⟶ FeCl₃

13. CO₂ + C ⟶ CO

14. KHCO₃ ⟶ K₂CO₃ + H₂O + CO₂

15. K + CO₂ ⟶ K₂O + C

16. Ca(OH)₂ + HNO₃ ⟶ Ca(NO₃)₂ + H₂O

17. K + H₂O ⟶ KOH + H₂

18. Ca + H₂O ⟶ Ca(OH)₂ + H₂

19. Al + H₂O ⟶ Al₂O₃ + H₂

20. Fe + H₂O ⇌ Fe₃O₄ + H₂

21. Zn + NaOH ⟶ Na₂ZnO₂ + H₂

22. Zn + HCl ⟶ ZnCl₂ + H₂

23. Al + H₂SO₄ ⟶ Al₂(SO₄)₃ + H₂

24. H₂ + O₂ ⟶ H₂O

25. N₂ + H₂ ⟶ NH₃

26. Fe₂O₃ + H₂O ⟶ Fe + H₂O

27. KBr + Cl₂ ⟶ KCl + Br₂

28. NaOH + Cl₂ ⟶ NaCl + NaClO + H₂O

29. NaHCO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + CO₂

30. Mg + CO₂ ⟶ MgO + C

31. Fe₂O₃ + CO ⟶ Fe + CO₂

32. CaO + HCl ⟶ CaCl₂ + H₂O

Answer

1. 2C + O₂ ⟶ 2CO

2. N₂ + O₂ ⇌ 2NO

3. 2ZnS + 3O₂ ⟶ 2ZnO + 2SO₂

4. 4Al + 3O₂ ⟶ 2Al₂O₃

5. 3Mg + N₂ ⟶ Mg₃N₂

6. 2Al + N₂ ⟶ 2AlN

7. 2NO + O₂ ⟶ 2NO₂

8. 2SO₂ + O₂ ⇌ 2SO₃

9. H₂ + Cl₂ ⟶ 2HCl

10. 2Fe + 3Cl₂ ⟶ 2FeCl₃

11. H₂S + Cl₂ ⟶ S + 2HCl

12. 2FeCl₂ + Cl₂ ⟶ 2FeCl₃

13. CO₂ + C ⟶ 2CO

14. 2KHCO₃ ⟶ K₂CO₃ + H₂O + CO₂

15. 4K + CO₂ ⟶ 2K₂O + C

16. Ca(OH)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O

17. 2K + 2H₂O ⟶ 2KOH + H₂

18. Ca + 2H₂O ⟶ Ca(OH)₂ + H₂

19. 2Al + 3H₂O ⟶ Al₂O₃ + 3H₂

20. 3Fe + 4H₂O ⟶ Fe₃O₄ + 4H₂

21. Zn + 2NaOH ⟶ Na₂ZnO₂ + H₂

22. Zn + 2HCl ⟶ ZnCl₂ + H₂

23. 2Al + 3H₂SO₄ ⟶ Al₂(SO₄)₃ + 3H₂

24. 2H₂ + O₂ ⟶ 2H₂O

25. N₂ + 3H₂ ⟶ 2NH₃

26. Fe₂O₃ + 3H₂ ⟶ 2Fe + 3H₂O

27. 2KBr + Cl₂ ⟶ 2KCl + Br₂

28. 2NaOH + Cl₂ ⟶ NaCl + NaClO + H₂O

29. 2NaHCO₃ + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O + 2CO₂

30. 2Mg + CO₂ ⟶ 2MgO + C

31. Fe₂O₃ + 3CO ⟶ 2Fe + 3CO₂

32. CaO + 2HCl ⟶ CaCl₂ + H₂O

Question 20

Write balanced equations for the following word equations :

1. Potassium nitrate ⟶ Potassium nitrite + Oxygen
2. Calcium + Water ⟶ Calcium hydroxide + Hydrogen
3. Iron + Hydrochloric acid ⟶ Iron [II] chloride + Hydrogen
4. Nitrogen dioxide + Water + Oxygen ⟶ Nitric acid
5. Lead dioxide [lead (IV) oxide] ⟶ Lead monoxide + Oxygen
6. Aluminium + Oxygen ⟶ Aluminium oxide
7. Iron + Chlorine ⟶ Iron [III] chloride
8. Potassium bromide + Chlorine ⟶ Potassium chloride + Bromine
9. Potassium bicarbonate ⟶ Potassium carbonate + Water + Carbon dioxide
10. Calcium hydroxide + Ammonium chloride ⟶ Calcium chloride + Water + Ammonia

Answer

1. 2KNO₃ ⟶ 2KNO₂ + O₂
2. Ca + 2H₂O ⟶ Ca(OH)₂ + H₂
3. Fe + 2HCl ⟶ FeCl₂ + H₂
4. 4NO₂ + 2H₂O + O₂ ⟶ 4HNO₃
5. 2PbO₂ ⟶ 2PbO + O₂
6. 4Al + 3O₂ ⟶ 2Al₂O₃
7. 2Fe + 3Cl₂ ⟶ 2FeCl₃
8. 2KBr + Cl₂ ⟶ 2KCl + Br₂
9. 2KHCO₃ ⟶ K₂CO₃ + H₂O + CO₂
10. Ca(OH)₂ + 2NH₄Cl ⟶ CaCl₂ + 2H₂O + 2NH₃

Question 21

Balance the following important equations :

1. NaHCO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + CO₂

2. NaOH + H₂SO₄ ⟶ Na₂SO₄ + H₂O

3. Pb(NO₃)₂ + NaCl ⟶ NaNO₃ + PbCl₂

4. FeSO₄ + NaOH ⟶ Na₂SO₄ + Fe(OH)₂

5. FeCl₃ + NaOH ⟶ NaCl + Fe(OH)₃

6. CuSO₄ + NaOH ⟶ Na₂SO₄ + Cu(OH)₂

7. FeCl₃ + NH₄OH ⟶ NH₄Cl + Fe(OH)₃

8. ZnO + NaOH ⟶ Na₂ZnO₂ + H₂O

9. Pb(OH)₂ + NaOH ⟶ Na₂PbO₂ + H₂O

10. Al₂O₃.2H₂O + NaOH ⟶ NaAlO₂ + H₂O

11. NaAlO₂ + H₂O ⟶ NaOH + Al(OH)₃

12. Al(OH)₃ ⟶ Al₂O₃ + H₂O

13. ZnS + O₂ ⟶ ZnO + SO₂

14. Fe₂O₃ + Al ⟶ Al₂O₃ + Fe

15. Al + Cl₂ ⟶ AlCl₃

16. NaCl + H₂SO₄ ⟶ Na₂SO₄ + HCl

17. Fe + HCl ⟶ FeCl₂ + H₂

18. Na₂CO₃ + HCl ⟶ NaCl + H₂O + CO₂

19. Pb(NO₃)₂ + HCl ⟶ PbCl₂ + HNO₃

20. AgCl + NH₄OH ⟶ Ag(NH₃)₂Cl + H₂O

21. MnO₂ + HCl ⟶ MnCl₂ + H₂O + Cl₂

22. Pb₃O₄ + HCl ⟶ PbCl₂ + H₂O + Cl₂

23. KMnO₄ + HCl ⟶ KCl + MnCl₂ + H₂O + Cl₂

24. K₂Cr₂O₇ + HCl ⟶ KCl + CrCl₃ + H₂O + Cl₂

25. NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + H₂O + NH₃

26. (NH₄)₂SO₄ + NaOH ⟶ Na₂SO₄ + H₂O + NH₃

27. Mg₃N₂ + H₂O ⟶ Mg(OH)₂ + NH₃

28. AlN + H₂O ⟶ Al(OH)₃ + NH₃

29. NH₃ + O₂ ⟶ N₂ + H₂O [burning of NH₃]

30. NH₃ + O₂ ⟶ NO + H₂O [Catalytic oxidation of NH₃]

31. NH₄OH + H₂SO₄ ⟶ (NH₄)₂SO₄ + H₂O

32. NH₃ + CuO ⟶ Cu + H₂O + N₂

33. NH₃ + Cl₂ ⟶ HCl + NCl₃ [nitrogen trichloride]

34. HNO₃ ⟶ H₂O + NO₂ + O₂

35. Ca(HCO₃)₂ + HNO₃ ⟶ Ca(NO₃)₂ + H₂O + CO₂

36. C + HNO₃ [conc.] ⟶ CO₂ + H₂O + NO₂

37. S + HNO₃ [conc.] ⟶ H₂SO₄ + H₂O + NO₂

38. Cu + HNO₃ [conc.] ⟶ Cu(NO₃)₂ + H₂O + NO₂

39. C + H₂SO₄ [conc.] ⟶ CO₂ + H₂O + SO₂

40. S + H₂SO₄ [conc.] ⟶ SO₂ + H₂O

41. Cu + H₂SO₄ [conc.] ⟶ CuSO₄ + H₂O + SO₂

Answer

1. 2NaHCO₃ + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O + 2CO₂

2. 2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

3. Pb(NO₃)₂ + 2NaCl ⟶ 2NaNO₃ + PbCl₂

4. FeSO₄ + 2NaOH ⟶ Na₂SO₄ + Fe(OH)₂

5. FeCl₃ + 3NaOH ⟶ 3NaCl + Fe(OH)₃

6. CuSO₄ + 2NaOH ⟶ Na₂SO₄ + Cu(OH)₂

7. FeCl₃ + 3NH₄OH ⟶ 3NH₄Cl + Fe(OH)₃

8. ZnO + 2NaOH ⟶ Na₂ZnO₂ + H₂O

9. Pb(OH)₂ + 2NaOH ⟶ Na₂PbO₂ + 2H₂O

10. Al₂O₃.2H₂O + 2NaOH ⟶ 2NaAlO₂ + 3H₂O

11. NaAlO₂ + 2H₂O ⟶ NaOH + Al(OH)₃

12. 2Al(OH)₃ ⟶ Al₂O₃ + 3H₂O

13. 2ZnS + 3O₂ ⟶ 2ZnO + 2SO₂

14. Fe₂O₃ + 2Al ⟶ Al₂O₃ + 2Fe

15. 2Al + 3Cl₂ ⟶ 2AlCl₃

16. 2NaCl + H₂SO₄ ⟶ Na₂SO₄ + 2HCl

17. Fe + 2HCl ⟶ FeCl₂ + H₂

18. Na₂CO₃ + 2HCl ⟶ 2NaCl + H₂O + CO₂

19. Pb(NO₃)₂ + 2HCl ⟶ PbCl₂ + 2HNO₃

20. AgCl + 2NH₄OH ⟶ Ag(NH₃)₂Cl + 2H₂O

21. MnO₂ + 4HCl ⟶ MnCl₂ + 2H₂O + Cl₂

22. Pb₃O₄ + 8HCl ⟶ 3PbCl₂ + 4H₂O + Cl₂

23. 2KMnO₄ + 8HCl ⟶ 2KCl + 2MnCl₂ + 4H₂O + Cl₂

24. K₂Cr₂O₇ + 14HCl ⟶ 2KCl + 2CrCl₃ + 7H₂O + 3Cl₂

25. 2NH₄Cl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O + 2NH₃

26. (NH₄)₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O + 2NH₃

27. Mg₃N₂ + 6H₂O ⟶ 3Mg(OH)₂ + 2NH₃

28. AlN + 3H₂O ⟶ Al(OH)₃ + NH₃

29. 4NH₃ + 3O₂ ⟶ 2N₂ + 6H₂O [burning of NH₃]

30. 4NH₃ + 5O₂ ⟶ 4NO + 6H₂O [Catalytic oxidation of NH₃]

31. 2NH₄OH + H₂SO₄ ⟶ (NH₄)₂SO₄ + 2H₂O

32. 2NH₃ + 3CuO ⟶ 3Cu + 3H₂O + N₂

33. NH₃ + 3Cl₂ ⟶ 3HCl + NCl₃ [nitrogen trichloride]

34. 4HNO₃ ⟶ 2H₂O + 4NO₂ + O₂

35. Ca(HCO₃)₂ + 2HNO₃ ⟶ Ca(NO₃)₂ + 2H₂O + 2CO₂

36. C + 4HNO₃ [conc.] ⟶ CO₂ + 2H₂O + 4NO₂

37. S + 6HNO₃ [conc.] ⟶ H₂SO₄ + 2H₂O + 6NO₂

38. Cu + 4HNO₃ [conc.] ⟶ Cu(NO₃)₂ + 2H₂O + 2NO₂

39. C + 2H₂SO₄ [conc.] ⟶ CO₂ + 2H₂O + 2SO₂

40. S + 2H₂SO₄ [conc.] ⟶ 3SO₂ + 2H₂O

41. Cu + 2H₂SO₄ [conc.] ⟶ CuSO₄ + 2H₂O + SO₂

Question 22

Give balanced equations for (1) & (2) by partial equation method, [steps are given below]

1. Reaction of excess ammonia with chlorine – Ammonia as a reducing agent
   (a) Ammonia first reacts with chlorine to give hydrogen chloride and nitrogen.
   (b) Hydrogen chloride then further reacts with excess ammonia to give ammonium chloride.
2. Oxidation of Lead [II] Sulphide by Ozone
   (a) Ozone first decomposes to give molecular oxygen & nascent oxygen.
   (b) Nascent oxygen then oxidizes lead [II] sulphide to lead [II] sulphate.

Answer

1. Reaction of excess ammonia with chlorine – Ammonia as a reducing agent

$\begin{array}{l} 2\text{NH}_3 + 3\text{Cl}_2 \longrightarrow \cancel{6\text{HCl}} + \text{N}_2 \\ 6\text{NH}_3 + \cancel{6\text{HCl}} \longrightarrow 6\text{NH}_4\text{Cl} \\ \hline 8\text{NH}_3 + 3\text{Cl}_2 \longrightarrow 6\text{NH}_4\text{Cl} + \text{N}_2 \end{array}$

2. Oxidation of Lead [II] Sulphide by Ozone

$\begin{array}{l} \text{O}_3 \longrightarrow \text{O}_2 + \cancel{\text{[O]}} \times 4 \\ \text{PbS} + \cancel{4\text{[O]}} \longrightarrow \text{PbSO}_4 \\ \hline \text{PbS} + 4\text{O}_3 \longrightarrow \text{PbSO}_4 + 4\text{O}_2 \end{array}$

Question 23

Define the terms – (a) Relative atomic mass (b) Relative molecular mass. State why indirect methods are utilized to determine the absolute mass of an atom. Explain in brief the indirect method used.

Answer

(a) The number of times one atom of an element is heavier than 1⁄12^th the mass of an atom of carbon [C¹²] is known as the Relative Atomic Mass [RAM] of the element.

$\begin{matrix} & \text{Mass of one atom} \\ \text{RAM} = & \text{of the element} \\ & \overline{(\frac{1}{12}) \space \text{Mass of one atom}} \\ & \text{of carbon [C}^{12}] \end{matrix}$

(b) The number of times one molecule of the substance is heavier than 1⁄12^th the mass of an atom of carbon [C¹²] is known as Relative Molecular Mass [RMM] of the element.

$\begin{matrix} & \text{Mass of one molecule} \\ \text{RMM} = & \text{of the substance} \\ & \overline{(\frac{1}{12}) \space \text{Mass of one atom}} \\ & \text{of carbon [C}^{12}] \end{matrix}$

As atoms are extremely small and very light, hence cannot be weighed directly. So indirect methods are utilized to determine the absolute mass of an atom.

The relative mass of an atom or molecule is hence considered by considering a mass of a light atom and relating the mass of other atoms or molecules to it.

Question 24(1)

Calculate relative molecular mass of

(a) ZnCO₃
(b) CaSO₄

[Zn = 65, S = 32, O = 16, Ca = 40, C = 12]

Answer

(a) Relative Molecular Mass of ZnCO₃:

Molecular wt. of ZnCO₃ = At. wt. of Zn + At. wt. of C + 3(At. wt. of O)
= 65 + 12 + 3(16)
= 65 + 12 + (48)
= 65 + 60
= 125

Hence, Relative Molecular Mass of ZnCO₃ = 125

(b) Relative Molecular Mass of CaSO₄:

Molecular wt. of CaSO₄ = At. wt. of Ca + At. wt. of S + 4(At. wt. of O)
= 40 + 32 + 4(16)
= 40 + 32 + 64
= 136

Hence, Relative Molecular Mass of CaSO₄ = 136

Question 24(2)

Calculate the percentage composition of

(a) calcium chloride
(b) calcium nitrate

[Ca = 40 , Cl = 35.5 , N = 14 , O = 16]

Answer

(a) Percentage composition of Calcium Chloride:

Chemical Formula of Calcium Chloride = CaCl₂

Mol. wt. of calcium chloride = At. wt. of Ca + 2(At. wt. of Cl)
= 40 + 2(35.5)
= 40 + 70
= 110

110 g of CaCl₂ contains 40 g of Calcium,

∴ 100 g of CaCl₂ will contain $\dfrac{40}{110}$ x 100 % = 36.36% of Calcium [percentage].

Similarly, 110 g of CaCl₂ will contain 70 g of Chlorine,

∴ 100 g of CaCl₂ will contain $\dfrac{70}{110}$ x 100 % = 63.64% of Chlorine.

Hence, Calcium Chloride contains 36.36% of Calcium and 63.64% of Chlorine.

(b) Percentage composition of Calcium Nitrate:

Chemical Formula of Calcium Nitrate = Ca(NO₃)₂

Mol. wt. of Calcium Nitrate = At. wt. of Ca + 2[At. wt. of N] + 6[At. wt. of O]
= 40 + 2(14) + 6(16)
= 40 + 28 + 96
= 164

164 g of Ca(NO₃)₂ contains 40 g of Calcium,

∴ 100 g of Ca(NO₃)₂ will contain $\dfrac{40}{164}$ x 100 % = 24.39% of Calcium.

Similarly, 164 g of Ca(NO₃)₂ will contain 28 g of Nitrogen,

∴ 100 g of Ca(NO₃)₂ will contain $\dfrac{28}{164}$ x 100 % = 17.07% of Nitrogen.

Similarly, 164 g of Ca(NO₃)₂ will contain 96 g of Oxygen,

∴ 100 g of Ca(NO₃)₂ will contain $\dfrac{96}{164}$ x 100 % = 58.54% of Oxygen.

Hence, Calcium Nitrate contains 24.39% of Calcium, 17.07% of Nitrogen and 58.54% of Oxygen.

Unit Test Paper 1 — The Lang of Chem

Question 1

Match the names of ions and radicals from 1 to 10 with their correct answer from A to Q.

Answer

Question 2

State which of the following formulas of compounds A to J are incorrect. If incorrect write the correct formula.

Answer

Question 3

Fill in the blanks with the correct word from the words in brackets :

1. A symbol represents a short form of a/an ............... [atom/element/molecule]
2. Compounds are always ............... [heterogeneous/homogeneous] in nature.
3. Variable valency is exhibited, since electrons are lost from an element from the ............... [valence/penultimate] shell.
4. A chemical equation is a shorthand form for a ............... [physical/chemical] change.
5. Relative molecular mass of an element/compound is the number of times one ............... of the substance is heavier than -1⁄12^th the mass of an atom of carbon [C¹²]. [atom/ion/molecule]

Answer

1. A symbol represents a short form of an element.
2. Compounds are always homogeneous in nature.
3. Variable valency is exhibited, since electrons are lost from an element from the penultimate shell.
4. A chemical equation is a shorthand form for a chemical change.
5. Relative molecular mass of an element/compound is the number of times one molecule of the substance is heavier than -1⁄12^th the mass of an atom of carbon [C¹²].

Question 4

Underline the compound in each equation given below, which is incorrectly balanced and write the correct balancing for the same.

1. Na₂SO₃ + HCl ⟶ 2NaCl + H₂O + SO₂
2. CaC₂ + N₂ ⟶ 2CaCN₂ + C
3. Fe₂O₃ + 2H₂ ⟶ 2Fe + 3H₂O
4. Cl₂ + 2H₂O + SO₂ ⟶ 4HCl + H₂SO₄
5. 6NaOH + 3Cl₂ ⟶ 6NaCl + NaClO₃ + 3H₂O
6. C₂H₅OH + 3O₂ ⟶ 2CO₂ + 2H₂O
7. NaOH + CO₂ ⟶ Na₂CO₃ + H₂O
8. 2H₂O + 2Cl₂ ⟶ 2HCl + O₂
9. 3CuO + NH₃ ⟶ 3Cu + 3H₂O + N₂
10. PbO₂ + 4HCl ⟶ PbCl₂ + H₂O + Cl₂

Answer

1. Na₂SO₃ + HCl ⟶ 2NaCl + H₂O + SO₂
   Corrected Balancing — 2HCl
2. CaC₂ + N₂ ⟶ 2CaCN₂ + C
   Corrected Balancing — CaCN₂
3. Fe₂O₃ + 2H₂ ⟶ 2Fe + 3H₂O
   Corrected Balancing — 3H₂
4. Cl₂ + 2H₂O + SO₂ ⟶ 4HCl + H₂SO₄
   Corrected Balancing — 2HCl
5. 6NaOH + 3Cl₂ ⟶ 6NaCl + NaClO₃ + 3H₂O
   Corrected Balancing — 5NaCl
6. C₂H₅OH + 3O₂ ⟶ 2CO₂ + 2H₂O
   Corrected Balancing — 3H₂O
7. NaOH + CO₂ ⟶ Na₂CO₃ + H₂O
   Corrected Balancing — 2NaOH
8. 2H₂O + 2Cl₂ ⟶ 2HCl + O₂
   Corrected Balancing — 4HCl
9. 3CuO + NH₃ ⟶ 3Cu + 3H₂O + N₂
   Corrected Balancing — 2NH₃
10. PbO₂ + 4HCl ⟶ PbCl₂ + H₂O + Cl₂
    Corrected Balancing — 2H₂O

Question 5

With reference to a chemical equation, state which of the statements 1 to 5 pertain to A or B.

A : Information provided by a chemical equation.
B : Limitations of a chemical equation

1. The nature of the individual elements.
2. The speed of the reaction.
3. The state of matter in which the substance is present.
4. The completion of the reaction.
5. The direction of the reaction.

Answer

1. A
2. B
3. A
4. B
5. A