Study of Compounds — Sulphuric Acid

Comment, sulphuric acid is referred to as:

(a) King of chemicals

(b) Oil of Vitriol

Answer

(a) Sulphuric acid is referred as the King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries. It has been known for a long time.

(b) Sulphuric acid is referred to as Oil of vitriol because in the later middle ages, it was obtained as an oily viscous liquid by heating crystals of green vitriol.

Sulphuric acid is manufactured by contact process.

(a) Give two balanced equations to obtain SO₂ in this process.

(b) Give the conditions for the oxidation of SO₂.

(c) Name the catalyst used.

(d) Why H₂SO₄ is not obtained by directly reacting SO₃ with water?

(e) Name the chemical used to dissolve SO₃ and also name the product formed. Give all the main reactions of this product.

Answer

(a) 4FeS₂ (iron pyrite) + 11O₂ ⟶ 2Fe₂O₃ + 8SO₂ ↑

S + O ⟶ SO₂ ↑

(b) The conditions for the oxidation of SO₂ are:

1. The temperature should be as low as possible. The yield has been found to be maximum at about 410 ⁰C – 450 ⁰C.
2. High pressure favours the reaction because the product formed has less volume than reactant. Pressure : 1 to 2 atmosphere.
3. Excess of oxygen increases the production of sulphur trioxide.

(c) Vanadium pentoxide [V₂O₅]

(d) Sulphuric acid is not obtained by directly reacting SO₃ with water since sulphur trioxide does not dissolve in water satisfactorily and it gives a lot of heat and forms misty droplets of sulphuric acid.

(e) The chemical used to dissolve SO₃ is concentrated sulphuric acid and the product formed is oleum.

Main reactions of this process are:

SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)

H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

Why the impurity of arsenic oxide must be removed before passing the mixture of SO₂ and air through the catalytic chamber?

Answer

Impurity of arsenic oxide must be removed before passing the mixture of SO₂ and air through the catalytic chamber otherwise the catalyst loses it's efficiency.

(a) Name the catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C.

(b) In the Contact process for the manufacture of sulphuric acid, sulphur trioxide is not converted to sulphuric acid by reacting it with water. Instead a two steps procedure is used. Write the equations for the two steps involved in D.

(c) What type of substance will liberate sulphur dioxide from sodium sulphite in step E?

(d) Write the equation for the reaction by which sulphur dioxide is converted to sodium sulphite in step F.

Answer

(a) Vanadium pentoxide

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons [450-500\text{\degree C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + \text{45 K cals}$

(b) Main reactions of this process are:

SO₃ + H₂SO₄ (conc.) ⟶ H₂S₂O₇ (oleum)

H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

(c) Dilute Sulphuric acid

Na₂SO₃ + H₂SO₄ ⟶ Na₂SO₄ + H₂O + SO₂[g] ↑

d) 2NaOH + SO₂ ⟶ Na₂SO₃ + H₂O

Why is water not added to concentrated H₂SO₄ in order to dilute it?

Answer

Water is never poured on concentrated H₂SO₄ to dilute it as large amount of heat is evolved which changes poured water to steam. The steam so formed causes spurting of acid which can cause burn injuries, so dilution is done by pouring acid on a given amount of water in a controlled manner by continuous stirring, else acid being heavier will settle down. The evolved heat is dissipated in the water itself and hence the spurting of the acid is minimized.

Give two balanced reactions of each type to show the following properties of sulphuric acid:

(a) Acidic nature

(b) Oxidising agent

(c) Dehydrating Nature

(d) Non-volatile nature

Answer

(a) Acidic nature

1. Mg + H₂SO₄ (dil.) ⟶ MgSO₄ + H₂ ↑
2. Zn + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂ ↑

(b) Oxidising agent

1. C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂ ↑
2. Zn + 2H₂SO₄ ⟶ ZnSO₄ + 2H₂O + SO₂ ↑

(c) Dehydrating Nature

1. $\text{HCOOH} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CO} + \text{H}_2\text{O}$
2. $\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} ↑ + \text{H}_2\text{O}$

(d) Non-volatile Nature

1. NaCl + H₂SO₄ ⟶ NaHSO₄ + HCl
2. KCl + H₂SO₄ ⟶ KHSO₄ + HCl

Give a chemical test to distinguish between:

(a) dilute sulphuric acid and dilute hydrochloric acid (using lead nitrate solution).

(b) dilute sulphuric acid and concentrated sulphuric acid.

Answer

(a) When dil sulphuric acid is reacted with lead nitrate it produces a white ppt. of lead sulphate which remains insoluble even on heating.

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

When dil hydrochloric acid is reacted with lead nitrate, a white ppt. of lead chloride is formed which dissolves on heating.

Pb(NO₃)₂ + 2HCl ⟶ 2HNO₃ + PbCl₂ ↓ [white ppt. formed]

(b) Dilute sulphuric acid when reacted with zinc produces Hydrogen gas which burns with a pop sound.

Zn + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂ ↑

Concentrated sulphuric acid when reacted with zinc produces SO₂ gas which turns acidified potassium dichromate paper green.

Zn + 2H₂SO₄ [conc.] ⟶ ZnSO₄ + 2H₂O + SO₂ ↑

Write balanced chemical equations: when hot and concentrated sulphuric acid reacts with the following:

(a) Sulphur

(b) NaOH

(c) Sugar

(d) Carbon

(e) Copper

Answer

(a) S + 2H₂SO₄ ⟶ 3SO₂ ↑ + 2H₂O

(b) 2NaOH + H₂SO₄ ⟶ Na₂SO₄ + 2H₂O

(c) C₁₂H₂₂O₁₁ $\xrightarrow {\text{Conc. H}_2\text{SO}_4}$ 12C + 11H₂O

(d) C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂ ↑

(e) Cu + 2H₂SO₄ ⟶ CuSO₄ + 2H₂O + SO₂ ↑

Why is:

(a) Concentrated sulphuric acid kept in air-tight bottles?

(b) H₂SO₄ is not used as a drying agent for H₂S?

(c) Sulphuric acid used in the preparation of HCl and HNO₃? Give equations in both cases.

Answer

(a) 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.

(b) Sulphuric acid is not a drying agent for H₂S because it reacts with H₂S to form sulphur.
H₂S + H₂SO₄ ⟶ S + 2H₂O + SO₂ ↑

(c) Concentrated sulphuric acid has a high boiling point (338°C) and so it is considered to be a non-volatile acid. It is, therefore, used for preparing volatile acids like hydrochloric acid and nitric acid from their salts by double decomposition.

NaCl + H₂SO₄ [conc.] $\xrightarrow{\lt 200 \degree\text{C}}$ NaHSO₄ + HCl

NaNO₃ + H₂SO₄ [conc.] $\xrightarrow{\lt 200 \degree\text{C}}$ NaHSO₄ + HNO₃

What property of conc. H₂SO₄ is made use of in each of the following cases? Give an equation for the reaction in each case:

(a) in the production of HCl gas when it reacts with a chloride,

(b) in the preparation of CO from HCOOH.

(c) as a source of hydrogen by diluting it and adding a strip of magnesium.

(d) in the preparation of sulphur dioxide by warming a mixture of conc. sulphuric acid and copper-turnings.

(e) hydrogen chloride gas is passed through concentrated sulphuric acid.

(f) it's reaction with (i) Ethanol (ii) Carbon.

Answer

(a) The property of being a non-volatile acid is used.

NaCl + H₂SO₄ [conc.] $\xrightarrow{\lt 200 \degree\text{C}}$ NaHSO₄ + HCl

(b) The property of being a dehydrating agent is used.

HCOOH $\xrightarrow{\text{Conc. H}_2\text{SO}_4}$ CO + H₂O

(c) The acidic property is used.

Mg + H₂SO₄ [dil.] ⟶ MgSO₄ + H₂ ↑

(d) The property of being an oxidizing agent is used.

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

(e) The property of being a drying agent is used.

Hydrogen chloride gas is dried by passing through conc. sulphuric acid as it does not react with conc. sulphuric acid.

(f) (i) The property of being a dehydrating agent is used.

$\underset{\text{ ethyl alcohol}}{\text{C}_2\text{H}_5\text{OH}} \xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4\text{[excess]}} \underset{ \text{ethene}}{\text{C}_2\text{H}_4} ↑ + \text{H}_2\text{O}\$

(ii) The property of being an oxidizing agent is used.

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

What is the name given to the salts of:

(a) Sulphurous acid

(b) Sulphuric acid

Answer

(a) Sulphite

(b) Sulphate

Give reasons for the following.

(a) Sulphuric acid forms two types of salts with NaOH.

(b) A piece of wood becomes black when concentrated sulphuric acid is poured on it.

(c) Brisk effervescence is seen when oil of vitriol is added to sodium carbonate.

Answer

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

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

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

(b) A piece of wood becomes black because of the dehydrating property of sulphuric acid. It readily removes elements of water from other compounds.

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

(c) When oil vitriol is added to sodium carbonate brisk effervescence are seen due to the evolution of Carbon dioxide.

Na₂CO₃ + H₂SO₄ ⟶ H₂O + Na₂SO₄ + CO₂ ↑

Copy and complete the following table:

Answer

Following are the typical properties of dilute acid. Complete them by inserting suitable words:

(i) Active metal + acid ⟶ ............... + ...............

(ii) Base + Acid ⟶ ............... + ...............

(iii) Carbonate/hydrogen carbonate + Acid ⟶ ............... + ...............

(iv) Sulphite/hydrogen sulphite + Acid ⟶ ............... + ...............

(v) Sulphide + Acid ⟶ ............... + ...............

Answer

(i) Active metal + acid ⟶ metallic salt + hydrogen gas

(ii) Base + Acid ⟶ salt + water

(iii) Carbonate/hydrogen carbonate + Acid ⟶ salt + water + carbon dioxide

(iv) Sulphite/hydrogen sulphite + Acid ⟶ salt + water + sulphur dioxide

(v) Sulphide + Acid ⟶ salt + hydrogen sulphide

(a) Which property of sulphuric acid accounts for it's use as a dehydrating agent?

(b) Concentrated sulphuric acid is both an oxidizing agent and a non-volatile acid. Write one equation each to illustrate the above mentioned properties of sulphuric acid.

Answer

(a) Sulphuric acid is a powerful dehydrating agent due to it's strong affinity towards water.

(b) (i) Oxidising agent : The oxidising property of concentrated sulphuric acid is due to the fact that on thermal decomposition it yields nascent oxygen [O].

H₂SO₄ ⟶ H₂O + SO₂ + [O]

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

C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂ ↑

(ii) Non-volatile nature : It has a high boiling point so it is considered to be a non-volatile acid. It is, therefore, used for preparing volatile acids like hydrochloric acid, nitric acid and acidic acid from their salts by double decomposition.

NaCl + H₂SO₄ ⟶ NaHSO₄ + HCl

Some properties of sulphuric acid are listed below. Choose the property A, B, C or D which is responsible for the reactions (i) to (v). Some properties may be repeated:

A. Typical acid

B. Dehydrating agent

C. Non-volatile acid

D. Oxidising 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

Answer

(i) Dehydrating agent

(ii) Oxidising agent

(iii) Non-volatile acid

(iv) Typical acid

(v) Typical acid

(a) Name the acid formed when sulphur dioxide dissolves in water.

(b) Name the gas released when sodium carbonate is added to a solution of sulphur dioxide.

Answer

(a) Sulphurous acid

(b) Carbon dioxide gas

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

(i) Copper nitrate

(ii) Zinc nitrate

(iii) Lead nitrate

(iv) Sodium nitrate

Answer

Lead nitrate

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

White ppt. of PbSO₄ is visible.

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

Answer

Liquid E is Ethanol [C₂H₅OH]

C₂H₅OH + H₂SO₄ (conc.) ⟶ C₂H₄ + H₂O

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

Answer

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

What property of conc. H₂SO₄

(i) is used in the action when sugar turns black in it's presence.

(ii) allows it to be used in the preparation of HCl and HNO₃ acids.

Answer

(i) Dehydrating agent.
C₁₂H₂₂O₁₁ + nH₂SO₄ ⟶ 12C + 11H₂O

(ii) Conc. H₂SO₄ is a non-volatile acid.

Write the equations for:

(i) dil. H₂SO₄ and barium chloride.

(ii) dil. H₂SO₄ and sodium sulphide.

Answer

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

(ii) Na₂S + H₂SO₄ ⟶ Na₂SO₄ + H₂S

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

Give the equation for:

(i) Heat on sulphur with conc. sulphuric acid

(ii) Reaction of - sugar with conc. sulphuric acid

Answer

(i) S + 2H₂SO₄ [conc.] ⟶ 3SO₂ + 2H₂O

(ii) C₁₂H₂₂O₁₁ + nH₂SO₄ ⟶ 12C + 11H₂O

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

Answer

ZnO + H₂SO₄ [dil.] ⟶ ZnSO₄ + H₂O

Select the correct answer 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

State your observation when – Sugar crystals are added to a hard glass test tube containing 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

Choose the correct answer from the choices:

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

(i) Hydrogen sulphide

(ii) Sulphur dioxide

(iii) Sulphur trioxide

(iv) Vapour of sulphuric acid.

Answer

Hydrogen sulphide

FeS + H₂SO₄ [dil.] ⟶ FeSO₄ + H₂S

Give a balanced equation for –

(i) Dilute sulphuric acid is poured over sodium sulphite

(ii) Manufacture of sulphuric acid by the Contact process.

Answer

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

(ii) Contact process:

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

Contact Tower :

$2\text{SO}_2 + \text{O}_2 \xrightleftharpoons[450-500\text{\degree C }/1 - 2 \text{ atmos}]{\text{V}_2\text{O}_5} 2\text{SO}_3 + 45 \text{ K cals}$

Absorption Tower :
SO₃ + H₂SO₄ ⟶ H₂S₂O₇

Dilution Tank:
H₂S₂O₇ + H₂O ⟶ 2H₂SO₄

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

(i) Potassium nitrate

(ii) Carbon

(iii) ethanol

Answer

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

$\text{KNO}_3 + \text{H}_2\text{SO}_4 [\text{conc.}]\xrightarrow{\lt 200 \degree\text{C}} \text{KHSO}_\text{4} + \text{HNO}_\bold{3}$

(ii) Sulphuric acid behaves as an oxidizing agent

C + 2H₂SO₄ ⟶ CO₂ + 2H₂O + 2SO₂

(iii) Dehydrating property

C₂H₅OH $\xrightarrow[170\degree\text{C}]{\text{Conc. H}_2\text{SO}_4}$ C₂H₄↑ + H₂O

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

Answer

Hydrogen sulphide (H₂S)

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

(i) $\text{CuSO}_4.5\text{H}_2\text{O} \xrightarrow{\text{Conc. H}_2 \text{SO}_4} \text{CuSO}_4 + 5\text{H}_2\text{O}$

(ii) S + 2H₂SO₄ [conc.] ⟶3SO₂ + 2H₂O

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

(iv) MgO + H₂SO₄ ⟶ MgSO₄ + H₂O

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

Answer

(i) B: Dehydrating agent

(ii) D: Oxidizing agent

(iii) C: Non-volatile acid

(iv) A: Dilute acid

(v) D: Oxidising agent

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

Answer

ZnS + H₂SO₄ (dil.) ⟶ ZnSO₄ + H₂S

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.

$\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

In the given equation :

S + 2H₂SO₄ ⟶ 3SO₂ + 2H₂O

Identify the role played by conc. H₂SO₄

(i) Non-volatile acid

(ii) Oxidising agent

(iii) Dehydrating agent

(iv) None of the above.

Answer

Oxidising agent

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

Answer

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

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]

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

Answer

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

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

State the conditions required for the following reactions to take place:

The conversion of sulphur dioxide to sulphur trioxide.

Answer

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

1. Temperature — 450 to 500°C
2. Pressure — 1 to 2 atmosphere
3. Presence of Catalyst — Vanadium pentoxide (V₂O₅)
4. Excess of Oxygen

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) $\text{CuSO}_4\text{.5H}_2\text{O} \xrightarrow{\text{Conc. H}_2\text{SO}_4} \text{CuSO}_4 \text{ [White Anhydrous]} + 5\text{H}_2\text{O}$

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

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

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

Give equations for the action of sulphuric acid on —

(i) Potassium hydrogen carbonate.

(ii) Sulphur

Answer

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

(ii) S + 2H₂SO₄ (conc.) ⟶ 3SO₂ + 2H₂O

Identify the acid in each case:

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

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

Answer

(i) Conc. sulphuric acid

(ii) Dilute sulphuric acid