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Chapter 7D

Study of Compounds — Sulphuric Acid

Class 10 - Dalal Simplified ICSE Chemistry Solutions


Equation Worksheet

Question 1

Sulphuric Acid — Manufacture-Contact process

Sulphuric AcidComplete and balance the equations
Manufacture-Contact process   
1. Step 1 [sulphur or pyrite burner]S + O2 ⟶ ............... ;
FeS2 + O2 ⟶ ............... + ...............
2. Step 2 [contact chamber]SO2 + O2 ⟶ ...............
3. Step 3 [absorbtion tower]SO3 + H2SO4 ⟶ ............... ;
............... + H2O ⟶ 2H2SO4

Answer

Sulphuric AcidComplete and balance the equations
Manufacture-Contact process   
1. Step 1 [sulphur or pyrite burner]S + O2SO2 ;
4FeS2 + 11O22Fe2O3 + 8SO2
2. Step 2 [contact chamber]2SO2 + O22SO3
3. Step 3 [absorbtion tower]SO3 + H2SO4H2S2O7 ;
H2S2O7 + H2O ⟶ 2H2SO4

Question 2

Chemical properties of sulphuric acid

Sulphuric AcidComplete and balance the equations
As an acid [dilute]   
4. Forms hydronium ions in aq. soln.H2SO4 + 2H2O ⟶ ............... + SO42-
Reaction with 
5. Active metal [zinc]Zn + H2SO4 ⟶ ............... + ............... [g]
6. Base [Sodium oxide]Na2O + H2SO4 ⟶ ............... + ...............
7. Base [Sodium hydroxide]NaOH + H2SO4 ⟶ ............... + ...............
8. Carbonate [Potassium carbonate]K2CO3 + H2SO4 ⟶ ............... + ............... + ............... [g]
9. Sulphite [Sodium sulphite]Na2SO3 + H2SO4 ⟶ ............... + ............... + ............... [g]
10. Bisulphite [Sodium bisulphite]NaHSO3 + H2SO4 ⟶ ............... + ............... + ............... [g]
11. Sulphide [Iron [II] sulphide]FeS + H2SO4 ⟶ ............... + ...............
As a dibasic acid 
12. Basicity is twoH2SO4 ⇌ ............... + SO42-
13. Dissociates in two stepsH2SO4 ⇌ ............... + ...............
HSO4- ⇌ ............... + SO42-
14. Forms two types of salts
- acid salt
- Normal salt
NaOH + H2SO4 ⟶ ............... + ...............
2NaOH + H2SO4 ⟶ ............... + ...............
As a non - volatile acid [conc.]
Displaces volatile acid from salt
 
15. Sodium chlorideNaCl + H2SO4 ⟶ ............... + ...............
16. Sodium nitrateNaNO3 + H2SO4 ⟶ ............... + ...............
As an oxidising agent [conc. acid]
Oxidation of
 
17. CarbonC + H2SO4 ⟶ ............... + H2O + ...............
18. SulphurS + H2SO4 ⟶ ............... + H2O
19. PhosphorousP + H2SO4 ⟶ ............... + H2O + ...............
20. CopperCu + H2SO4 ⟶ ............... + H2O + ...............
21. ZincZn + H2SO4 ⟶ ............... + H2O + ...............
22. Hydrogen iodideHI + H2SO4 ⟶ ............... + H2O + ...............
23. Hydrogen sulphideH2S + H2SO4 ⟶ ............... + H2O + ...............
As a dehydrating agent [conc. acid] 
24. GlucoseC6H12O6 ⟶ ............... + H2O
25. SucroseC12H22O11 ⟶ ............... + H2O
26. Cellulose[C6H10O5]n ⟶ ............... + H2O
27. Formic acidH.COOH ⟶ ............... + H2O
28. Oxalic acidH2C2O4 ⟶ ............... + ............... + H2O
29. Hydrated copper sulphateCuSO4.5H2O ⟶ ............... + H2O

Answer

Sulphuric AcidComplete and balance the equations
As an acid [dilute]   
4. Forms hydronium ions in aq. soln.H2SO4 + 2H2O ⟶ 2H3O+ + SO42-
Reaction with 
5. Active metal [zinc]Zn + H2SO4ZnSO4 + H2 [g]
6. Base [Sodium oxide]Na2O + H2SO4Na2SO4 + H2O
7. Base [Sodium hydroxide]2NaOH + H2SO4Na2SO4 + 2H2O
8. Carbonate [Potassium carbonate]K2CO3 + H2SO4K2SO4 + H2O + CO2 [g]
9. Sulphite [Sodium sulphite]Na2SO3 + H2SO4Na2SO4 + H2O + SO2 [g]
10. Bisulphite [Sodium bisulphite]2NaHSO3 + H2SO4Na2SO4 + 2H2O + 2SO2 [g]
11. Sulphide [Iron [II] sulphide]FeS + H2SO4FeSO4 + H2S
As a dibasic acid 
12. Basicity is twoH2SO42H+ + SO42-
13. Dissociates in two stepsH2SO4H+ + HSO4-
HSO4-H+ + SO42-
14. Forms two types of salts
- acid salt
- Normal salt
NaOH (insufficient) + H2SO4NaHSO4 + H2O
2NaOH (excess) + H2SO4Na2SO4 + 2H2O
As a non - volatile acid [conc.]
Displaces volatile acid from salt
 
15. Sodium chlorideNaCl + H2SO4 [conc.] ⟶ NaHSO4 + HCl (<200°C)
16. Sodium nitrateNaNO3 + H2SO4 [conc.] ⟶ NaHSO4 + HNO3 (<200°C)
As an oxidising agent [conc. acid]
Oxidation of
 
17. CarbonC + 2H2SO4 [conc.] ⟶ CO2 + 2H2O + 2SO2
18. SulphurS + 2H2SO4 [conc.] ⟶ 3SO2 + 2H2O
19. Phosphorous2P + 5H2SO4 [conc.] ⟶ 2H3PO4 + 2H2O + 5SO2
20. CopperCu + 2H2SO4 [conc.] ⟶ CuSO4 + 2H2O + SO2
21. ZincZn + 2H2SO4 [conc.] ⟶ ZnSO4 + 2H2O + SO2
22. Hydrogen iodide2HI + H2SO4 [conc.] ⟶ I2 + 2H2O + SO2
23. Hydrogen sulphideH2S + H2SO4 [conc.] ⟶ S + 2H2O + SO2
As a dehydrating agent [conc. acid] 
24. GlucoseC6H12O6 + H2SO4 [conc.] ⟶ 6C + 6H2O
25. SucroseC12H22O11 + H2SO4 [conc.] ⟶ 12C + 11H2O
26. Cellulose[C6H10O5]n + H2SO4 [conc.] ⟶ 6[C]n + 5[H2O]n
27. Formic acidH.COOH + H2SO4 [conc.] ⟶ CO + H2O
28. Oxalic acidH2C2O4 + H2SO4 [conc.] ⟶ CO + CO2 + H2O
29. Hydrated copper sulphateCuSO4.5H2O + H2SO4 [conc.] ⟶ CuSO4 + 5H2O

Questions

Question 1(2002)

State the substance/s reacted with dilute or concentrated sulphuric acid to form the following gases:

(i) Hydrogen

(ii) Carbon dioxide.

State whether the acid used in each case is dilute or concentrated.

Answer

(i) Zinc, Iron or any active metal reacts with dilute Sulphuric Acid (H2SO4) to form Hydrogen. The reactions are shown below:

Zn + H2SO4 (dil.) ⟶ ZnSO4 + H2 [g]

Fe + H2SO4 (dil.) ⟶ FeSO4 + H2 [g]

(ii) Carbonates and bicarbonates react with dilute Sulphuric Acid (H2SO4) to give Carbon dioxide (CO2). The reactions are shown below:

Na2CO3 + H2SO4 (dil.) ⟶ Na2SO4 + H2O + CO2

2KHCO3 + H2SO4 (dil.) ⟶ K2SO4 + 2H2O + 2CO2

Question 2(2002)

Write the equations for the lab. preparation of:

(i) Na2SO4 using dil. H2SO4,

(ii) PbSO4 using dil. H2SO4

Answer

(i) 2NaOH + H2SO4 (dil) ⟶ Na2SO4 + 2H2O

(ii) Pb(NO3)2 + H2SO4 (dil) ⟶ PbSO4 ↓ + 2HNO3

Question 1(2003)

State the name of the process by which H2SO4 is manufactured. Name the catalyst used.

Answer

Contact process is the name of the process by which H2SO4 is manufactured.

Catalyst — Vanadium pentoxide [V2O5] or platinum [Pt.]

Question 2(2003)

Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is ............... [less volatile / stronger] in comparison to these two acids.

Answer

Concentrated sulphuric acid is used in the laboratory preparation of nitric acid and hydrochloric acid because it is less volatile in comparison to these two acids.

Question 3(2003)

Write the equations for the laboratory preparation of the following salts using sulphuric acid:

(i) Copper sulphate from copper

(ii) Lead sulphate from lead nitrate

Answer

Cu + 2H2SO4 (conc.) ⟶ CuSO4 + SO2 + 2H2O

Pb(NO3)2 + H2SO4 (dil.) ⟶ PbSO4 ↓ + 2HNO3

Question 1(2004)

Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide.

Answer

Vanadium pentoxide [V2O5] or platinum [Pt.]

Question 2(2004)

In the Contact process, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two- step procedure is used. Write the equations for the two steps involved.

Answer

The equations for the two steps involved are:

SO3 + H2SO4 ⟶ H2S2O7

H2S2O7 + H2O ⟶ 2H2SO4

Question 1(2005)

Write balanced equations for the following:

(i) Potassium hydrogen carbonate and dilute sulphuric acid.

(ii) Sodium nitrate and concentrated sulphuric acid.

Answer

2KHCO3 + H2SO4 ⟶ K2SO4 + 2H2O + 2CO2

NaNO3+H2SO4[conc.]<200°CNaHSO4+HNO3\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}

Question 2(2005)

Choose the property of sulphuric acid (A, B, C or D), which is relevant to each of the preparations (i) to (iii) :

A: Dil. acid (typical acid properties)

B: Non-volatile acid

C: Oxidizing agent

D: Dehydrating agent

Preparation of :

(i) HCl

(ii) ethene from ethanol

(iii) copper sulphate from copper oxide.

Answer

(i) In preparation of HCl — B: Non-volatile acid

NaCl+H2SO4[conc.]<200°CNaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}

(ii) In preparation of ethene from ethanol — D: Dehydrating agent

C2H5OH + H2SO4 (conc.) ⟶ C2H4 + H2O

(iii) In preparation of copper sulphate from copper oxide — A: Dil. acid (typical acid properties)

CuO + H2SO4 (dil.) ⟶ CuSO4 + H2O

Question 1(2006)

Name the process used for the large scale manufacture of sulphuric acid.

Answer

Contact process.

Question 2(2006)

Which property of sulphuric acid accounts for it's use as a dehydrating agent.

Answer

Sulphuric acid removes water of crystallization from hydrated salts.

Question 3(2006)

Conc. H2SO4 is an oxidizing agent and a non volatile acid. Write an equation for each property.

Answer

Sulphuric acid as an Oxidizing agent —

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

Sulphuric acid as an non-volatile acid —

NaNO3+H2SO4[conc.]<200°CNaHSO4+HNO3\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_3

Question 1(2007)

Write balanced equation for the following reactions:

(i) Lead sulphate from lead nitrate solution and dilute sulphuric acid.

(ii) Copper sulphate from copper and conc. sulphuric acid.

Answer

(i) Pb(NO3)2 + H2SO4 [dil] ⟶ PbSO4 + 2HNO3

(ii) Cu + 2H2SO4 [conc.] ⟶ CuSO4 + 2H2O + SO2

Question 2(2007)

Properties of H2SO4 are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v).

A : Acid
B: Dehydrating agent
C: Non-volatile acid
D: Oxidizing agent

(i) C12H22O11+ nH2SO4 ⟶ 12C + 11H2O + nH2SO4

(ii) S + 2H2SO4 ⟶ 3SO2 + 2H2O

(iii) NaCl + H2SO4 ⟶ NaHSO4 + HCl

(iv) CuO + H2SO4 ⟶ CuSO4 + H2O

(v) Na2CO3 + H2SO4 ⟶Na2SO4 + H2O + CO2

(Some properties may be repeated)

Answer

(i) B: Dehydrating agent

(ii) D: Oxidizing agent

(iii) C: Non-volatile acid

(iv) A : Acid

(v) A : Acid

Question 3(2007)

Dilute hydrochloric acid and dilute sulphuric acid are both colourless solutions. How will the addition of barium chloride solution to each help to distinguish between the two.

Answer

Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO4) but with dil. HCl no white ppt. is produced.

H2SO4 (dil.) + BaCl2 (aq.) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

HCl (dil.) + BaCl2 (aq.) ⟶ no reaction

Question 4(2007)

From HCl, HNO3, H2SO4 state which has the highest boiling point and which has the lowest.

Answer

H2SO4 — highest boiling point (338°C).

HCl — lowest boiling point (-83°C).

Question 1(2008)

Dilute H2SO4 will produce a white ppt. when added to a solution of:

A. Copper nitrate
B. Zinc nitrate
C. Lead nitrate
D. Sodium nitrate

Answer

Lead nitrate

Pb(NO3)2 + H2SO4 (dil) ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed]

White ppt. of PbSO4 is visible.

Question 2(2008)

Identify the following substance : Liquid E can be dehydrated to product ethene.

Answer

Liquid E is Ethanol [C2H5OH]

C2H5OH + H2SO4 (conc.) ⟶ C2H4 + H2O

Question 3(2008)

Copy and complete the following table relating to an important industrial process and it's final output.

Name of processInputsCatalystEquation for catalyzed reactionOutput
Contact ProcessSulphur dioxide + oxygen      

Answer

Name of processInputsCatalystEquation for catalyzed reactionOutput
Contact ProcessSulphur dioxide + oxygenVanadium pentoxide [V2O5] or platinum [Pt.]2SO2+O2450500°C /12 atmosV2O52SO3+45 K cals2\text{SO}_2 + \text{O}_2 \xrightleftharpoons [450-500\text{\degree C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \text{45 K cals}Sulphuric acid

Question 4(2008)

Making use only of substances given: dil. sulphuric acid, Sodium carbonate, Zinc, Sodium Sulphite, Lead, Calcium carbonate:

Give equations for the reactions by which you could obtain:

(i) hydrogen

(ii) sulphur dioxide

(iii) carbon dioxide

(iv) zinc carbonate (2 steps)

Answer

(i) Zn + H2SO4 [dil.] ⟶ ZnSO4 + H2

(ii) Na2SO3 + H2SO4 [dil.] ⟶ Na2SO4 + H2O + SO2

(iii) Na2CO3 + H2SO4 [dil.] ⟶ Na2SO4 + H2O + CO2

(iv) Zn + H2SO4 [dil.] ⟶ ZnSO4 + H2
ZnSO4 + Na2CO3 [dil.] ⟶ ZnCO3 + Na2SO4

Question 5(2008)

What property of conc. H2SO4 is used in the action when sugar turns black in it's presence.

Answer

Dehydrating agent.

C12H22O11+ nH2SO4 ⟶ 12C + 11H2O

Question 6(2008)

Write the equations for:

(1) dil. H2SO4 and barium chloride.

(2) dil. H2SO4 and sodium sulphide.

Answer

(i) BaCl2 + H2SO4 (dil) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

(ii) Na2S + H2SO4 ⟶ Na2SO4 + H2S

Question 7(2008)

Which property of conc. H2SO4 allows it to be used in the preparation of HCl and HNO3 acids.

Answer

Conc. H2SO4 is a non volatile acid.

Question 1(2009)

Name the gas evolved [formula is not acceptable] – The gas that can be oxidized to sulphur.

Answer

Hydrogen sulphide gas (H2S)

H2S + H2SO4 [conc.] ⟶ S + SO2 + 2H2O

Question 1(2010)

Give the equation for:

(i) Heat on sulphur with conc. H2SO4

(ii) Reaction of - sugar with conc. H2SO4

Answer

(i) S + 2H2SO4 [conc.] ⟶ 3SO2 + 2H2O

(ii) C12H22O11+ nH2SO4 ⟶ 12C + 11H2O

Question 2(2010)

Give a balanced equation for the conversion of zinc oxide to zinc sulphate.

Answer

ZnO + H2SO4 [dil.] ⟶ ZnSO4 + H2O

Question 3(2010)

Select from A, B, C –

A: Sodium hydroxide solution.

B: A weak acid.

C: Dilute sulphuric acid.

The solution which liberates sulphur dioxide gas, from sodium sulphite.

Answer

Dilute sulphuric acid

Na2SO3 + H2SO4 [dil.] ⟶ Na2SO4 + H2O + SO2

Question 1(2011)

State your observation when – Sugar crystals are added to conc. sulphuric acid.

Answer

Steam is evolved from the test tube and it becomes very hot. Black spongy mass of carbon is formed.

C12H22O11+ nH2SO4 ⟶ 12C + 11H2O

Question 2(2011)

Choose the correct answer from the choices:

The gas evolved when dil. sulphuric acid reacts with iron sulphide.

A: Hydrogen sulphide
B: Sulphur dioxide
C: Sulphur trioxide
D: Vapour of sulphuric acid.

Answer

Hydrogen sulphide

FeS + H2SO4 [dil.] ⟶ FeSO4 + H2S

Question 3(2011)

Give a balanced equation for – Dilute sulphuric acid is poured over sodium sulphite

Answer

Na2SO3 + H2SO4 (dil.) ⟶ Na2SO4 + H2O + SO2

Question 4(2011)

Give balanced equations for the manufacture of sulphuric acid by the Contact process.

Answer

Contact process:

Sulphur or Pyrite Burner:
S + O2 ⟶ SO2
4FeS2 + 11O2 ⟶ 2Fe2O3 + 8SO2

Contact Tower :

2SO2+O2450500°C /12 atmosV2O52SO3+45 K cals2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[450-500\text{\degree C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + 45 \text{ K cals}

Absorption Tower :
SO3 + H2SO4 ⟶ H2S2O7

Dilution Tank:
H2S2O7 + H2O ⟶ 2H2SO4

Question 5(2011)

State the property of sulphuric acid shown by the reaction of conc. sulphuric acid when heated with

(a) Potassium nitrate

(b) Carbon

Answer

(a) Sulphuric acid behaves as a non volatile acid.

KNO3+H2SO4[conc.]<200°CKHSO4+HNO3\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{KHSO}_\text{4} + \text{HNO}_\bold{3}

(b) Sulphuric acid behaves as an oxidizing agent

C + 2H2SO4 ⟶ CO2 + 2H2O + 2SO2

Question 1(2012)

Name the gas produced on reaction of dilute sulphuric acid with a metallic sulphide.

Answer

Hydrogen sulphide (H2S)

Question 2(2012)

Some properties of sulphuric acid are listed below. Choose the role played by sulphuric acid – A, B, C, or D which is responsible for the reactions (i) to (v).

Some role/s may be repeated.

A : Dilute acid
B : Dehydrating agent
C : Non-volatile acid
D : Oxidising agent

(1) CuSO4.5H2OConc. H2SO4CuSO4+5H2O\text{CuSO}_4.5\text{H}_2\text{O} \xrightarrow{\text{Conc. H}_2 \text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O}

(2) S + 2H2SO4 [conc.] ⟶3SO2 + 2H2O

(3) NaNO3+H2SO4[conc.]<200°CNaHSO4+HNO3\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}

(4) MgO + H2SO4 ⟶ MgSO4 + H2O

(5) Zn + 2H2SO4 [conc.] ⟶ ZnSO4 + SO2 + 2H2O

Answer

(1) B: Dehydrating agent

(2) D: Oxidizing agent

(3) C: Non-volatile acid

(4) A: Dilute acid

(5) D: Oxidising agent

Question 3(2012)

Give balanced equation for the reaction : Zinc sulphide and dilute sulphuric acid.

Answer

ZnS + H2SO4 (dil.) ⟶ ZnSO4 + H2S

Question 1(2013)

State one appropriate observation for : conc. H2SO4 is added to a crystal of hydrated copper sulphate.

Answer

The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

CuSO4.5H2OConc. H2SO4CuSO4 [White Anhydrous]+5H2O\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}

Question 2(2013)

In the given equation :

S + 2H2SO4 ⟶ 3SO2 + 2H2O

Identify the role played by conc. H2SO4 i.e.

(A) Non-volatile acid

(B) Oxidising agent

(C) Dehydrating agent

(D) None of the above.

Answer

Oxidizing agent

Question 3(2013)

Give a balanced equation for : Dehydration of concen­trated sulphuric acid with sugar crystals.

Answer

C12H22O11Conc. H2SO412C [sugar charcoal]+11H2O\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}

Question 4(2013)

Identify the substance underlined: A dilute mineral acid which forms a white precipitate when treated with barium chloride solution.

Answer

Dilute sulphuric acid.

BaCl2 + H2SO4 (dil) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

Question 1(2014)

Write balanced equations for the following: Action of concentrated sulphuric acid on carbon.

Answer

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

Question 2(2014)

Distinguish between the following pairs of compounds using the test given within brackets: Dilute sulphuric acid and dilute hydrochloric acid [using barium chloride solution]

Answer

Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO4) but with dilute hydrochloric acid no white ppt. is produced since barium chloride is soluble in dil. sulphuric acid.

H2SO4 (dil.) + BaCl2 (aq.) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

HCl (dil.) + BaCl2 (aq.) ⟶ No reaction

Question 3(2014)

State – Any two conditions for the conversion of sulphur dioxide to sulphur trioxide.

Answer

Two conditions for the conversion of sulphur dioxide to sulphur trioxide are:

  1. Temperature — 450 to 500°C
  2. Pressure — 1 to 2 atmosphere

Question 4(2014)

Give one equation each to show the following properties of sulphuric acid:

(i) Dehydrating property

(ii) Acidic nature

(iii) As a non-volatile acid

Answer

(i) CuSO4.5H2OConc. H2SO4CuSO4 [White Anhydrous]+5H2O\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}

(ii) Na2SO3 + H2SO4 (dil.) ⟶ Na2SO4 + H2O + SO2

(iii) NaNO3+H2SO4[conc.]<200°CNaHSO4+HNO3\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_\bold{3}

Question 1(2015)

Identify the acid in each case:

(i) The acid which is used in the preparation of a non­-volatile acid.

(ii) The acid which produces sugar charcoal from sugar.

(iii) The acid on mixing with lead nitrate soln. produces a white ppt. which is insoluble even on heating.

Answer

(i) Conc. nitric acid

(ii) Conc. sulphuric acid

(iii) Dilute sulphuric acid

Question 2(2015)

Give equations for the action of sulphuric acid on —

(a) Potassium hydrogen carbonate.

(b) Sulphur

Answer

(a) 2KHCO3 + H2SO4 ⟶ K2SO4 + 2H2O + 2CO2

(b) S + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O

Question 3(2015)

In the manufacture of sulphuric acid by the Contact process, give the equations for the conversion of sulphur trioxide to sulphuric acid.

Answer

SO3 + H2SO4 ⟶ H2S2O7

H2S2O7 + H2O ⟶ 2H2SO4

Question 1(2016)

Write balanced chemical equation for: Action of dilute sulphuric acid on Sodium Sulphite.

Answer

Na2SO3 + H2SO4(dil) ⟶ Na2SO4 + H2O + SO2

Question 2(2016)

State your observations when:

(i) Barium chloride soln. is mixed with sodium sulphate soln.

(ii) Concentrated sulphuric acid is added to sugar crystals.

Answer

(i) White coloured precipitate of barium sulphate is formed when sodium sulphate is mixed with barium chloride.

Na2SO4 (dil.) + BaCl2 (aq.) ⟶ 2NaCl + BaSO4 ↓ [white ppt. formed]

(ii) Black spongy charred mass of carbon is formed. Steam is seen to be evolved.

C12H22O11Conc. H2SO412C [sugar charcoal]+11H2O\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}

Question 3(2016)

A, B, C and D summarize the properties of sulphuric acid depending on whether it is dilute or concentrated.

A: Typical acid property

B: Non-volatile acid

C: Oxidizing agent

D: Dehydrating agent

Choose the property [A, B, C or D] depending on which is relevant to each of the following:

(i) Preparation of hydrogen chloride gas.

(ii) Preparation of copper sulphate from copper oxide.

(iii) Action of conc. sulphuric acid on sulphur.

Answer

(i) B: Non-volatile acid

NaCl+H2SO4[conc.]<200°CNaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}

(ii) A: Typical acid property

CuO + H2SO4 (conc.) ⟶ CuSO4 + H2O

(iii) C: Oxidizing agent

S + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O

Question 1(2017)

Write the balanced chemical equation for – Action of concentrated sulphuric acid on sulphur.

Answer

S + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O

Question 2(2017)

State one relevant observation for – Action of conc. sulphuric acid on hydrated copper sulphate.

Answer

The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

CuSO4.5H2OConc. H2SO4CuSO4 [White Anhydrous]+5H2O\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}

Question 3(2017)

State – How will you distinguish between dilute hydrochloric acid and dilute sulphuric acid using lead nitrate solution.

Answer

Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.

H2SO4 (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed which does not dissolve on warming the mixture].

Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.

2HCl (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbCl2 ↓ [white ppt. formed which dissolves on warming the mixture].

Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

Question 4(2017)

Write balanced chemical equations to show –

(i) The oxidizing action of conc. sulphuric acid on carbon.

(ii) The behaviour of H2SO4 as an acid when it reacts with magnesium.

(iii) The dehydrating property of conc. sulphuric acid with sugar.

(iv) The conversion of SO3 to sulphuric acid in the Contact process.

Answer

(i) C + 2H2SO4 (conc.) ⟶ CO2 + 2SO2 + 2H2O

(ii) Mg + H2SO4 (dil.) ⟶ MgSO4 + H2 (g)

(iii) C12H22O11Conc. H2SO412C [sugar charcoal]+11H2O\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}

(iv) SO3 + H2SO4 (conc.) ⟶ H2S2O7 (oleum)
H2S2O7 + H2O ⟶ 2H2SO4 (conc.)

Question 1(2018)

Choose the correct answer from the options given:

The catalyst used in the Contact Process is:

  1. Copper
  2. Iron
  3. Vanadium pentoxide
  4. Manganese dioxide

Answer

Vanadium pentoxide

Question 2(2018)

Write a balanced chemical equation for - Action of concentrated sulphuric acid on carbon.

Answer

C + 2H2SO4 (conc.) ⟶ CO2 + 2SO2 + 2H2O

Question 3(2018)

State which property of sulphuric acid is shown by the reaction of concentrated sulphuric acid with:

(i) Ethanol

(ii) Carbon

Answer

(i) Dehydrating property

C2H5OHConc. H2SO4C2H4 [etheneethylene]+H2O\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}

(ii) Oxidizing property

C + 2H2SO4 (conc.) ⟶ CO2 + 2SO2 + 2H2O

Question 1(2019)

Write balanced chemical equations for the following reactions:

Action of dilute sulphuric acid on —

(i) Sodium hydroxide

(ii) Zinc sulphide

Answer

(i) 2NaOH + H2SO4 ⟶ Na2SO4 + 2H2O

(ii) ZnS + H2SO4 (dil.) ⟶ ZnSO4 + H2S

Question 2(2019)

Distinguish between the following pairs of compounds using the reagent given in the bracket:

Dilute hydrochloric acid and dilute sulphuric acid [using lead nitrate soln.]

Answer

Sulphuric acid forms a white precipitate with lead nitrate solution. This precipitate does not dissolve on warming the reaction mixture.

H2SO4 (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed which does not dissolve on warming the mixture].

Hydrochloric acid forms a white precipitate with lead nitrate solution. This precipitate dissolves on warming the reaction mixture so as to form clear solution.

2HCl (dil.) + Pb(NO3)2 ⟶ 2HNO3 + PbCl2 ↓ [white ppt. formed which dissolves on warming the mixture].

Hence, dilute hydrochloric acid and dilute sulphuric acid can be distinguished using lead nitrate solution.

Question 3(2019)

Complete the following equation:

C + conc. H2SO4

Answer

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

Question 4(2019)

Copy and complete the following table which refers to the industrial method for preparation of sulphuric acid.

Name of the compoundName of the processCatalytic equation [with the catalyst]
Sulphuric acid

Answer

Name of the compoundName of the processCatalytic equation [with the catalyst]
Sulphuric acidContact Process2SO2+O2450500°C /12 atmosV2O52SO3+45 K cals2\text{SO}_2 + \text{O}_2 \xrightleftharpoons [450-500\text{\degree C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \text{45 K cals}

Question 1(2020)

Choose the correct answer from the options:

The acid which can produce carbon from cane sugar, is —

  1. Concentrated Hydrochloric acid
  2. Concentrated Nitric acid
  3. Concentrated Sulphuric acid
  4. Concentrated Acetic acid

Answer

Concentrated Sulphuric acid

C12H22O11Conc. H2SO412C [sugar charcoal]+11H2O\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}

Question 2(2020)

Identify the substance underlined in the following :

The acid that is a dehydrating as well as a drying agent.

Answer

Sulphuric acid

Additional Questions

Question 1

State why sulphuric acid was called – 'oil of vitriol'.

Answer

Sulphuric acid was initially called 'oil of vitriol' because it was prepared by distilling green vitriol [FeSO4.7H2O] and hence the name – 'oil of vitriol' was given.

2FeSO4.7H2O ⟶ Fe2O3 + SO2 + SO3 + 14H2O

SO3 + 14H2O ⟶ H2SO4 + 13H2O

Question 2

State how you would convert (i) sulphur (ii) chlorine (iii) sulphur dioxide — to sulphuric acid.

Answer

(i) By action of heat on sulphur and nitric acid:
S + 6HNO3 [conc.] ⟶ 6NO2 + 2H2O + H2SO4

(ii) By passage of chlorine through an aqueous solution of sulphur dioxide:
Cl2 + SO2 + 2H2O ⟶ 2HCl + H2SO4

(iii) By oxidation of an aq. soln. of sulphur dioxide:
2SO2 + 2H2O + O2 ⟶ 2H2SO4

Question 3

State the purpose of the 'Contact process'.

Answer

The purpose of the 'Contact process' is to manufacture sulphuric acid.

Question 4

In the Contact process

(i) State how you would convert (a) sulphur (b) iron pyrites to sulphur dioxide in the first step of the Contact process.

(ii) State the conditions i.e. catalyst, promoter, temperature and pressure in the catalytic oxidation of sulphur dioxide to sulphur trioxide in the Contact tower. Give a balanced equation for the same.

(iii) State why the above catalytic oxidation reaction supplies energy.

(iv) Give a reason why – vanadium pentoxide is preferred to platinum during the catalytic oxidation of sulphur dioxide.

(v) Give a reason why the catalyst mass is heated electrically – only initially.

(vi) State why sulphur trioxide vapours are absorbed in concentrated sulphuric acid and not in water to obtain sulphuric acid.

Answer

(i) Sulphur dioxide is produced by burning sulphur [S] or iron pyrites [FeS2]

S + O2 ⟶ SO2

4FeS2 + 11O2 ⟶ 2Fe2O3 + 8SO2

(ii) Catalyst : Vanadium pentoxide [V2O5] or platinum [Pt.]

Temperature : 450 - 500°C , Pressure : 1 to 2 atmosphere

2SO2+O2450500°C/12 atmosV2O52SO3+Δ2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[450-500\text{\degree C} /1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \Delta

Conversion ratio: 98% of sulphur dioxide converted to sulphur trioxide.

(iii) The catalytic oxidation of SO2 to SO3 is an exothermic reaction, hence, supplies energy.

(iv) Vanadium pentoxide is preferred to platinized asbestos as a catalyst since it is comparatively cheaper and less easily poisoned or susceptible to impurities.

(v) The catalyst mass is only initially heated electrically, since the catalytic oxidation of sulphur dioxide is an exothermic reaction and the heat produced maintains the temperature at 450 – 500°C.

(vi) Even though sulphur trioxide is an acid anhydride of sulphuric acid it is not directly absorbed in water to give sulphuric acid.

The reaction is highly exothermic resulting in production of a dense fog of sulphuric acid particles which do not condense easily.

Hence sulphur trioxide vapours are dissolved in conc. sulphuric acid to give oleum which on dilution with the requisite amount of soft water in the dilution tank gives sulphuric acid of the desired concentration [about 98%].

Question 5

Give a reason why concentrated sulphuric acid is kept in air tight bottles.

Answer

As, concentrated sulphuric acid is hygroscopic and has a great affinity for water hence, it absorbs moisture from the atmosphere. Therefore, it is kept in air tight bottles.

Question 6

State the basic steps with reasons, involved in diluting a beaker of conc. H2SO4.

Answer

For dilution of conc. H2SO4, the conc. acid is added to water and not the water to the acid even though heat is evolved in both cases.

  1. If water is added to the acid : There is a sudden increase in the temperature and the acid being in bulk tends to spurt out.
  2. If acid is added to water : The water is in bulk and the acid being heavier settles down. The heat evolved is dissipated in the water itself and hence the spurting of the acid is minimized.

Question 7

Give reasons why dilute sulphuric acid:

(i) behaves as an acid when dilute.

(ii) is dibasic in nature.

Answer

(i) Acidic properties of sulphuric acid are due to the presence of hydronium ions [H3O+] formed when H2SO4 dissociates in aqueous solution. Hence, it behaves as an acid when dilute.

(ii) Sulphuric acid dissociates in aqueous solution giving two hydrogen ions (H+) ions per molecule of the acid. Hence, it is dibasic in nature.

H2SO4 ⇌ 2H+ + SO42-

H2SO4 + 2H2O ⇌ 2H3O+ + SO42-

Question 8

Convert dil. H2SO4 to –

(i) Hydrogen

(ii) Carbon dioxide

(iii) Sulphur dioxide

(iv) Hydrogen sulphide

(v) An acid salt

(vi) A normal salt.

Answer

(i) Zn + H2SO4 (dil.) ⟶ ZnSO4 + H2 (g)

(ii) 2KHCO3 + H2SO4 ⟶ K2SO4 + 2H2O + 2CO2

(iii) S + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O

(iv) FeS + H2SO4 (dil.) ⟶ FeSO4 + H2S

(v) NaOH + H2SO4 (dil.)⟶ NaHSO4 + H2O

(vi) 2NaOH + H2SO4 (dil.)⟶ Na2SO4 + 2H2O

Question 9

Give equations for formation of two different acids from conc. H2SO4. State the property of sulphuric acid involved in the above formation.

Answer

NaNO3+H2SO4[conc.]<200°CNaHSO4+HNO3\text{NaNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HNO}_3

NaCl+H2SO4[conc.]<200°CNaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}

The property of being a non-volatile acid is used in the above reactions.

Question 10

Give equations for oxidation of conc. H2SO4 giving the oxidised products –

(i) Carbon dioxide

(ii) Sulphur dioxide

(iii) Phosphoric acid

(iv) Copper (II) sulphate

(v) Iodine

(vi) Sulphur

respectively

Answer

(i) C + 2H2SO4 (conc.) ⟶ CO2 + 2SO2 + 2H2O

(ii) S + 2H2SO4 (conc.) ⟶ 3SO2 + 2H2O

(iii) 2P + 5H2SO4 (conc.) ⟶ 2H3PO4 + 5SO2 + 2H2O

(iv) Cu + 2H2SO4 (conc.) ⟶ CuSO4 + SO2 + 2H2O

(v) 2HI + H2SO4 (conc.) ⟶ I2 + SO2 + 2H2O

(vi) H2S + H2SO4 (conc.) ⟶ S + SO2 + 2H2O

Question 11

Give a reason why concentrated and not dil H2SO4 behaves as an oxidising and dehydrating agent.

Answer

Oxidising property of concentrated sulphuric acid is due to the fact that on thermal decomposition, it yields nascent oxygen [O] which helps in oxidation.

H2SO4 ⟶ H2O + SO2 + [O]

Nascent oxygen oxidizes non-metals, metals and inorganic compounds.

On the other hand, dil H2SO4 on heating does not decompose to give nascent oxygen and as such cannot behave as an Oxidising agent.

Question 12

Give the equation for the reaction of conc. sulphuric acid with –

(i) glucose

(ii) sucrose

(iii) cellulose

(iv) an organic acid containing one carbon atom and two hydrogen atoms

(v) an organic acid containing two carbon and two hydrogen atoms

(vi) an alcohol

(vii) hydrated copper [II] sulphate

Answer

(i) Reaction of conc. sulphuric acid with glucose:

C6H12O6Conc. H2SO46C+6H2O\text{C}_{6}\text{H}_{12}\text{O}_{6}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6C} + \text{6H}_2\text{O}

(ii) Reaction of conc. sulphuric acid with sucrose:

C12H22O11Conc. H2SO412C [sugar charcoal]+11H2O\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}

(iii) Reaction of conc. sulphuric acid with cellulose:

[C6H10O5]nConc. H2SO46[C]n+5[H2O]n[\text{C}_{6}\text{H}_{10}\text{O}_{5}]\text{n}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6[C]}_\text{n} + \text{5[H}_2\text{O}]_\text{n}

(iv) Reaction of conc. sulphuric acid with formic acid [H.COOH]:

H.COOHConc. H2SO4CO+H2O\text{H.COOH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CO} + \text{H}_2\text{O}

(v) Reaction of conc. sulphuric acid with oxalic acid [H2C2O4]:

H2C2O4Conc. H2SO4CO+CO2+H2O\text{H}_{2}\text{C}_{2}\text{O}_{4} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CO} + \text{CO}_{2} + \text{H}_2\text{O}

(vi) Reaction of conc. sulphuric acid with ethyl alcohol:

C2H5OHConc. H2SO4C2H4 [etheneethylene]+H2O\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}

(vii) Reaction of conc. sulphuric acid with hydrated copper sulphate:

CuSO4.5H2OConc. H2SO4CuSO4 [White Anhydrous]+5H2O\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}

Question 13

State the observation seen when conc. H2SO4 is added to –

(i) sucrose

(ii) hydrated copper [II] sulphate.

Answer

(i) Conc. H2SO4 dehydrates sucrose to a black spongy charged mass of carbon sugar charcoal.

C12H22O11Conc. H2SO412C [sugar charcoal]+11H2O\text{C}_{12}\text{H}_{22}\text{O}_{11}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{12C} \small{\space[sugar\space charcoal]} + \text{11H}_2\text{O}

(ii) The blue coloured hydrous copper sulphate changes to white anhydrous copper sulphate as the water of crystallization is removed.

CuSO4.5H2OConc. H2SO4CuSO4 [White Anhydrous]+5H2O\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}

Question 14

State how addition of –

(i) copper

(ii) NaCl

to hot conc. H2SO4 serves as a test for the latter.

Answer

(i) When Cu is added to hot conc. H2SO4, colourless suffocating (SO2) gas is evolved. The gas turns orange coloured acidified potassium dichromate solution to green colour.

Cu + 2H2SO4 (conc.) Δ\xrightarrow{\Delta} CuSO4 + SO2 + 2H2O

(ii) When NaCl is heated with conc. H2SO4, colourless pungent smelling HCl gas is evolved.

NaCl+H2SO4[conc.]<200°CNaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}

Dense white fumes are produced when a rod dipped in NH3 soln. is brought near the test tube containing HCl gas.

Question 15

Give two tests for dilute sulphuric acid with balanced equations. State why

(i) BaCl2

(ii) Pb(NO3)2

are used for the above tests

Answer

(i) BaCl2 solution on treating with dilute sulphuric acid forms white ppt. of barium sulphate, which is insoluble in dilute sulphuric acid.

BaCl2 + H2SO4 (dil) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

(ii) Pb(NO3)2 on treating with dilute sulphuric acid gives white ppt of lead sulphate, which is insoluble in dil. H2SO4.

Pb(NO3)2 + H2SO4 (dil) ⟶ 2HNO3 + PbSO4 ↓ [white ppt. formed]

BaCl2 and Pb(NO3)2 are used for the above tests as they react with H2SO4 and form BaSO4 and PbSO4 respectively, which are the only compounds insoluble in dil. sulphuric acid.

Question 16

Give a test to distinguish dilute sulphuric acid from dilute HCl and dilute HNO3.

Answer

Barium chloride (BaCl2) solution reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO4) but with dil. HCl and dil. HNO3 no white ppt. is produced since BaCl2 and Ba(NO3)2 is soluble in dil. H2SO4.

H2SO4 (dil.) + BaCl2 (aq.) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

HNO3 (dil.) + BaCl2 (aq.) ⟶ No ppt.

HCl (dil.) + BaCl2 (aq.) ⟶ No ppt.

Question 17

State three different chemical compounds other than acids manufactured industrially from sulphuric acid.

Answer

  1. Ammonium sulphate [(NH4)2SO4] used as a fertilizer
  2. Superphosphate of lime [Ca(H2PO4)2 + (CaSO4)] used as a fertilizer
  3. Tri-nitro toluene (T.N.T.) used as an explosive.

Unit Test Paper 7D — Sulphuric Acid

Question 1

Select the correct answer from the choice in brackets.

  1. The oxidised product obtained when sulphur reacts with conc. H2SO4. [H2S/SO2/H2SO3].
  2. The dehydrated product obtained when cane sugar reacts with conc. H2SO4. [CO/C/CO2]
  3. The type of salt formed when excess of caustic soda reacts with sulphuric acid. [acid salt / normal salt]
  4. The reduced product obtained when hydrogen sulphide reacts with conc. H2SO4. [SO2/S/H2O]
  5. The salt which reacts with dil. H2SO4 acid to give an insoluble ppt. [Cu(NO3)2/Zn(NO3)2/Pb(NO3)2]

Answer

  1. SO2
  2. C
  3. Normal salt
  4. SO2
  5. Pb(NO3)2

Question 2

Iron pyritesAColourless acidic gasBSulphur trioxideCOleumDSulphuric acid\text{Iron pyrites} \xrightarrow{\text{A}} \text{Colourless acidic gas} \xrightarrow{\text{B}} \text{Sulphur trioxide} \xrightarrow{\text{C}} \text{Oleum} \xrightarrow{\text{D}} \text{Sulphuric acid}

  1. Give a balanced equation for the conversion 'A'.
  2. The gaseous mixture of the product of conversion 'A' and air contains dust particles as an impurity. Name another impurity in the same mixture.
  3. Is the conversion 'B' an exothermic or an endothermic reaction. Would lowering the temperature favour or retard the forward reaction.
  4. If the product of conversion 'B' is an acid anhydride of H2SO4, the anhydride of conversion 'A' is ............... .
  5. State why water is added for the conversion 'D' and not for the conversion 'C'

Answer

  1. 4FeS2 + 11O2 ⟶ 2Fe2O3 + 8SO2
  2. Arsenious oxide (As4O6)
  3. The catalytic oxidation of SO2 to SO3 is an exothermic reaction. According to Le Chatelier's principle, higher yield of the product is obtained by lowering the temperature.
  4. acid anhydride of H2SO3 (Sulphurous acid)
  5. Sulphur trioxide is not directly absorbed in water to give sulphuric acid as the reaction is highly exothermic resulting in production of a dense fog of sulphuric acid particles which do not condense easily.
    Hence, sulphur trioxide vapours are dissolved in conc. sulphuric acid to give oleum which on dilution with the requisite amount of soft water in the dilution tank gives sulphuric acid of the desired concentration.

Question 3

Give balanced equations for the following reactions using sulphuric acid.

  1. Formation of a black mark on a piece of wood on addition of conc. H2SO4 to it.
  2. Oxidation of a foul smelling acidic gas, heavier than air and fairly soluble in H2O by conc. H2SO4.
  3. Formation of an acid salt from sulphuric acid and (a) an alkali (b) a sodium salt.
  4. Formation of a hydrocarbon from an organic compound
  5. Formation of sulphur dioxide using a metal below hydrogen in the activity series.

Answer

  1. [C6H10O5]nConc. H2SO46[C]n+5[H2O]n[\text{C}_{6}\text{H}_{10}\text{O}_{5}]\text{n}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{6[C]}_\text{n} + \text{5[H}_2\text{O}]_\text{n}

  2. H2S + H2SO4 (conc.) ⟶ S + 2H2O + SO2

  3. (a) Formation of an acid salt from sulphuric acid and an alkali (NaOH):
    NaOH [insufficient] + H2SO4 (dil.)⟶ NaHSO4 + H2O

    (b) Formation of an acid salt from sulphuric acid and a sodium salt (NaCl):
    NaCl+H2SO4[conc.]<200°CNaHSO4+HCl\text{NaCl} + \text{H}_2\text{SO}_4 [\text{conc.}] \xrightarrow{\lt 200 \degree\text{C}} \text{NaHSO}_\text{4} + \text{HCl}

  4. C2H5OHConc. H2SO4C2H4 [etheneethylene]+H2O\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}

  5. Cu + 2H2SO4 (conc.) ⟶ CuSO4 + 2H2O +SO2

Question 4

Match the conversions in column 'X' using sulphuric acid, with the type of chemical property of sulphuric acid A to E it represents in column 'Y'

XY
1. Nitre → Nitric acidA: As an oxidising agent
2. Copper [II] oxide → Copper [II] sulphateB: As a dibasic acid
3. Copper → Copper [II] sulphateC: As an acid when dilute
4. Ethanol [ethyl alcohol] → EtheneD: As a least or non-volatile acid
5. Sodium hydroxide → Sodium bisulphate and sodium sulphateAs a dehydrating agent.

Answer

XY
1. Nitre → Nitric acidD: As a least or non-volatile acid
2. Copper [II] oxide → Copper [II] sulphateC: As an acid when dilute
3. Copper → Copper [II] sulphateA: As an oxidising agent
4. Ethanol [ethyl alcohol] → EtheneE: As a dehydrating agent.
5. Sodium hydroxide → Sodium bisulphate and sodium sulphateB: As a dibasic acid

Question 5

Select the correct substance from the substances A to J which react with the sulphuric acid to give the product 1 to 10. [State whether the acid used in each case is dilute or concentrated].

A : Iron

B : Sodium carbonate

C : Sodium chloride

D : Formic acid

E : Sodium nitrate

F : Sodium sulphite

G : Ethyl alcohol

H : Sodium sulphide

I : Sodium hydroxide (excess)

J : Hydrogen sulphide

  1. Product — Sulphur dioxide
  2. Product — Sulphur
  3. Product — Hydrogen
  4. Product — Hydrochloric acid
  5. Product — Sodium sulphate
  6. Product — Carbon dioxide
  7. Product — Carbon monoxide
  8. Product — Nitric acid
  9. Product — Hydrogen sulphide
  10. Product — Ethene

Answer

  1. F: Sodium sulphite and dil. sulphuric acid
  2. J: Hydrogen sulphide and conc. sulphuric acid
  3. A: Iron and dil. sulphuric acid
  4. C: Sodium chloride and conc. sulphuric acid
  5. I: Sodium hydroxide and dil. sulphuric acid
  6. B: Sodium carbonate and dil. sulphuric acid
  7. D: Formic acid and conc. sulphuric acid
  8. E: Sodium nitrate and conc. sulphuric acid
  9. H: Sodium sulphide and dil. sulphuric acid
  10. G: Ethyl alcohol and conc. sulphuric acid

Question 6.1

Give reasons for the following:

Sulphuric acid forms two types of salts with an alkali.

Answer

Sulphuric acid forms two types of salts i.e., acid salt and normal salt with an alkali since it's basicity is two.

NaOH [insufficient] + H2SO4 ⟶ H2O + NaHSO4 [acid salt]

2NaOH [excess] + H2SO4 ⟶ 2H2O + Na2SO4 [normal salt]

Question 6.2

Give reasons for the following:

Conc. sulphuric acid is used as a laboratory reagent in the preparation of iodine from hydrogen iodide.

Answer

As conc. H2SO4 oxidises hydrogen iodide to iodine, hence, conc. sulphuric acid is used as a laboratory reagent in the preparation of iodine from hydrogen iodide.

Question 6.3

Give reasons for the following:

Barium chloride solution can be used to distinguish between dil. H2SO4 and dil HNO3.

Answer

Barium chloride soln. reacts with dilute sulphuric acid to give a white ppt. of barium sulphate (BaSO4) but with dil. HNO3 no white ppt. is produced since Ba(NO3)2 is soluble in dil. H2SO4.

H2SO4 (dil.) + BaCl2 (aq.) ⟶ 2HCl + BaSO4 ↓ [white ppt. formed]

HNO3 (dil.) + BaCl2 (aq.) ⟶ No ppt.

Question 6.4

Give reasons for the following:

The gaseous product obtained differs when zinc reacts with dilute and with conc. H2SO4 respectively.

Answer

Concentrated Sulphuric acid is a strong oxidising agent because on thermal decomposition it yields nascent oxygen which helps in oxidation. Hence, when zinc reacts with conc. H2SO4, Zn is oxidized to ZnSO4 and sulphur dioxide gas is evolved along with hydrogen gas.

Zn + 2H2SO4 (conc.) ⟶ ZnSO4 + 2H2 + 2SO2

Dilute Sulphuric acid on the other hand behaves as a typical acid. Hence, when zinc reacts with dil. H2SO4, displacement reaction takes place, ZnSO4 is formed and hydrogen gas is evolved.

Zn + H2SO4 (dil.) ⟶ ZnSO4 + H2

Question 6.5

Give reasons for the following:

Ethanol can be converted to ethene using sulphuric acid.

Answer

As H2SO4 is a strong dehydrating agent, so it removes elements of water i.e. hydrogen and oxygen in the ratio 2:1 from carbohydrates, organic compounds etc. Hence, ethanol can be converted to ethene using sulphuric acid.

C2H5OHConc. H2SO4C2H4 [etheneethylene]+H2O\text{C}_{2}\text{H}_{5}\text{OH}\xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{C}_{2}\text{H}_{4} \small{\space[ethene-ethylene]} + \text{H}_2\text{O}

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