Class 10 SELINA Solutions Chemistry Chapter 11 - Study of Compounds D. Sulphuric Acid

Intext 1

Ex. 11

Study of Compounds D. Sulphuric Acid Exercise Intext 1

Solution 1

(a) Sulphuric acid is called King of Chemicals because there is no other manufactured compound which is used by such a large number of key industries.

(b) Sulphuric acid is referred to as Oil of vitriol as it was obtained as an oily viscous liquid by heating crystals of green vitriol.

Solution 2

(a) Two balanced equations to obtain SO₂ is:

(i) 4FeS₂ + 11O₂ 2Fe₂O₃ +8SO₂

(ii) S +O₂ SO₂

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

(i) The temperature should be as low as possible. The yield has been found to be maximum at about 410⁰C-450^oC

(ii) High pressure (2 atm) is favoured because the product formed has less volume than reactant.

(iii) Excess of oxygen increases the production of sulphur trioxide.

(iv) Vanadium pentoxide or platinised asbestos is used as catalyst.

(d) Sulphuric acid is not obtained directly by reacting SO₃ with water because the reaction is highly exothermic which produce the fine misty droplets of sulphuric acid that is not directly absorbed by water.

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

Reaction involved in this process:

SO₃ + H₂SO₄ → H₂S₂O₇

Solution 3

Impurity of ARSENIC poisons the catalyst [i.e. deactivates the catalyst]. So, it must be removed before passing the mixture of SO₂ air through the catalytic chamber.

Solution 4

(a) The catalyst which helps in the conversion of sulphur dioxide to sulphur trioxide in step C is Vanadium pentoxide.

(b) The two steps for the conversion of sulphur trioxide to sulphuric acid is:

(i) SO₃+ H₂SO₄H₂S₂O₇

(ii) H₂S₂O₇ + H₂O 2H₂SO₄

(c)The substance that will liberate sulphur dioxide in step E is dilute H₂SO_4.

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

SO₂+2NaOHNa₂SO₃+H₂O

Na₂O+SO₂Na₂SO₃

Study of Compounds D. Sulphuric Acid Exercise Ex. 11

Solution B 1

(a) Active metal + Acid ⟶ Metal sulphate + Hydrogen

(b) Base + Acid ⟶ Salt + Water

(d) Sulphide/hydrogen sulphite + Acid ⟶ Salt + Water + Sulphur dioxide

(e) Sulphide + Acid ⟶ Salt + Hydrogen sulphide

Solution B 2(a)

The acid formed when sulphur dioxide dissolves in water is sulphurous acid.

Solution B 2(b)

Carbondioxide gas is released when sodium carbonate is added to solution of sulphur dioxide.

Solution B 2(e)

Dilute sulphuric acid.

Solution B 3

The name of the salt of

(a) Hydrogen sulphites and Sulphites.

(b) Sulphate and bisulphate.

Solution B 6

(i) B

(ii) C

(iii) A

(iv) A

Solution B 9

Column 1 Substance reacted with acid	Column 2 Dilute or concentrated acid	Column 3 Gas
Zinc	Dilute sulphuric acid	Hydrogen
Calcium carbonate	Concentrated sulphuric acid	Carbon dioxide
Bleaching power CaOCl₂	Dilute sulphuric acid	Only chlorine

Solution C 1

(a) Two types of salts are formed when sulphuric acid reacts with NaOH because sulphuric acid is dibasic.

NaOH + H₂SO₄NaHSO₄ + H₂O

2NaOH + H₂SO₄Na₂SO₄ + 2H₂O

(b) A piece of wood becomes black when concentrated sulphuric acid is poured on it because it gives a mass of carbon.

Na₂CO₃+H₂SO₄Na₂SO₄+H₂O+CO₂

Solution C 2

Water is not added to concentrated acid since it is an exothermic reaction. If water is added to the acid, there is a sudden increase in temperature and the acid being in bulk tends to spurt out with serious consequences.

Solution C 3

(a) Concentrated sulphuric acid is hygroscopic substance that absorbs moisture when exposed to air. Hence, it is stored in air tight bottles.

(b) Sulphuric acid is not a drying agent for H₂S because it reacts with H₂S to form sulphur.

H₂SO₄+H₂S 2H₂O+SO₂+S

(c) Concentrated sulphuric acid has high boiling point (356^oC). So, it is considered to be non-volatile. Hence, it is used for preparing volatile acids like Hydrochloric acid and Nitric acids from their salts by double decomposition.

NaCl+H₂SO₄NaHSO₄ +HCl

NaNO₃ +H₂SO₄NaHSO₄ +HNO₃

Solution C 5

(a)

(b)

(d) C₁₂H₂₂O₁₁ + Conc. H₂SO₄ → 6C + 6H₂O

(e) C₁₂H₂₂O₁₁+ Conc. H₂SO₄→ 6C + 6H₂O

(f) C + H₂SO₄ → CO₂ + 2H₂O + 2SO₂.

Solution C 6

(a) BaCl₂ + H₂SO₄ → BaSO₄↓ + 2HCl

dil.

(b) Na₂S + H₂SO₄→ Na₂SO₄ + H₂S↑

dil.

Solution C 9(c)

Sulphuric acid precipitates the insoluble sulphate of barium from the solution of barium chloride.

BaCl₂ + H₂SO₄→ BaSO₄ + 2HCl

Dilute HCl does not react with barium chloride solution, and thus, no precipitate is produced in the reaction.

Solution C 11

(a) Zn + dil. H₂SO₄→ ZnSO₄ + H₂

(b) Na₂CO₃ + dil. H₂SO₄→ Na₂SO₄ + H₂O + CO₂

(d) Zn + dil. H₂SO₄→ ZnSO₄ + H₂

ZnSO₄ + Na₂CO₃→ ZnCO₃ + Na₂SO₄

Solution C 13

In the contact process for the manufacture of sulphuric acid, the equations for the conversion of sulphur trioxide to sulphuric acid are

SO₃+ H₂SO₄→ H₂S₂O₇

(oleum or pyrosulphuric acid)

H₂S₂O₇+ H₂O → 2H₂SO₄

Solution D 1

(a) Due to its reducing property. i.e, it is a non-volatile acid.

NaCl+ H₂SO₄NaHSO₄ + HCl

(Conc.)

(b) It is a dehydrating agent.

HCOOH CO + H₂O

(c) Acidic property: Magnesium is present above hydrogen in the reactivity series so sulphuric acid is able to liberate hydrogen gas by reacting with magnesium strip.

Mg + H₂SO₄MgSO₄+H₂

(d) Due to its oxidizing character

Cu +H₂SO₄CuSO₄ +2H₂O +SO₂

(e) Due to its dehydrating property Hydrogen chloride gas is passed through concentrated sulphuric acid, it dries hydrogen chloride gas.

H₂SO₄ + H₂O → H₃O⁺ + HSO₄⁻

(f)

(i) Reaction with Ethanol - dehydrating agent

(ii) Reaction with Carbon - oxidising agent

Solution D 2(a)

Sulphuric acid is powerful dehydrating agent on account of its strong affinity towards water.

Solution D 2(b)

Concentrated sulphuric acid as

i. Oxidising agent:

The oxidising property of conc. sulphuric acid its due to the fact that on thermal decomposition, it yeilds nacent oxygen [O].

H₂SO₄→ H₂O + SO₂+ [O]

ii. Non-volatile acid:

conc. sulphuric acid has high boiling point (338°C) that why it is said to be a non volitile compound, therefore it is used for preparing volatile acids like hydrochloric acids, nitric acids from there salts by double decomposition

H₂SO₄ + NaCl → NaHSO₄ + HCl

Solution 2

Balanced reactions are:

(a) Acidic nature:

(i) Dilute H₂SO₄ reacts with basic oxides to form sulphate and water.

2 NaOH+H₂SO₄ Na₂SO₄+2H₂O

(ii) CuO+H₂SO₄CuSO₄+H₂O

(iii) It reacts with carbonate to produce CO₂.

Na₂CO₃+H₂SO₄Na₂SO₄+H₂O+CO₂

(b) Oxidising agent:

H₂SO₄ H₂O +SO₂ +[O]

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

For example,

Carbon to carbon dioxide

C+H₂SO₄CO₂ +H₂O +2SO₂

Sulphur to sulphur dioxide

S +H₂SO₄3SO₂ +2H₂O

(d) Non-volatile nature:

It has a high boiling point (356^oC) so it is considered to be non-volatile. Therefore, it is used for preparing volatile acids like hydrochloric acid, nitric acid from their salts by double decomposition reaction.

NaCl + H₂SO₄NaHSO₄ +HCl

KCl + H₂SO₄KHSO₄ +HCl

Solution 3

(a) When dilute hydrochloric acid is treated with lead nitrate, lead chloride formed is soluble in hot water and when dilute sulphuric acid is treated with lead nitrate lead sulphate formed does not dissolve in hot water.

(b)

1. Dilute sulphuric acid treated with zinc gives Hydrogen gas which bums with pop sound.

Concentrated H₂SO₄ gives SO₂ gas with zinc and the gas turns Acidified potassium dichromate paper green.

2.Barium chloride solution gives white ppt. with dilute H₂SO₄, This white ppt. is insoluble in all acids.

Concentrated H₂SO₄ and NaCl mixture when heated gives dense white fumes if glass rod dipped in Ammonia solution is brought near it.

Solution 4

(a) When sulphuric acid reacts with sulphur the product formed is Sulphur dioxide is formed.

S +2H₂SO₄3SO₂ + 2H₂O

(b) When sulphuric acid reacts with sodium hydroxide it neutralizes base to form sodium sulphate.

2NaOH + H₂SO₄Na₂SO₄ + 2H₂O

C₁₂ H₂₂O₁₁12C + 11H₂O

(d) When sulphuric acid reacts with carbon it forms carbon dioxide and sulphur dioxide gas.

C +2H₂SO₄CO₂ + 2H₂O + 2SO₂

(e) When sulphuric acid reacts with copper it forms copper sulphate and sulphur dioxide.

Cu + H₂SO₄CuSO₄ + 2H₂O + SO₂

Solution A 1

(a) Hydrogen sulphide

Solution A 2

(b) Oxidising agent

Solution A 3

Solution B 2(c)

Liquid E is Ethanol.

Solution B 2(d)

Hydrogen Chloride.

Solution B 2(f)

Hydrogen Sulphide (H₂S).

Solution B 2(g)

Sulphuric acid (H₂SO₄)

Solution B 2(h)

Concentrated sulphuric acid

Solution B 2(i)

Concentrated sulphuric acid.

Solution B 4

(a) The property of concentrated sulphuric acid which allows it to be used in is used in the action when sugar turns black in its presence is its dehydrating property.

(b) The property of concentrated sulphuric acid which allows it to be used in the preparation of hydrogen chloride and nitric acid is its non-volatility.

H₂SO₄+ BaCl₂ → BaSO₄+ 2HCl

Solution B 5

(a) Non-volatile nature

(b) as an oxidising agent

Solution B 7

(a) (B) Dehydrating agent

(b) (D) Oxidising agent

(d) (A) Dilute acid

(e) (D) Oxidising agent

Solution B 8

Name of process

Inputs

Catalyst

Equation for catalyzed reaction output

output

Contact process

Sulphur dioxide + oxygen

Platinum or V₂O₅

2SO₂ + O₂ ⇄ 2SO₃

Sulphuric acid

Solution C 4(a)

Charring of sugar takes place. Sulphuric acid dehydrates sugar leaving behind carbon which is black in colour.

Solution C 4(b)

When Conc. H₂SO₄ is added to a crystal of hydrated copper sulphate,it removes water of crystalisation from salt.

Solution C 10

Two conditions for the conversion of sulphur dioxide to sulphur trioxide is as follows:

Temperature: 450-500° C Catalyst: V₂O₅

Solution C 12

(a) Action of sulphuric acid on potassium hydrogen carbonate

2KHCO₃+ H₂SO₄→ K₂SO₄+ 2H₂O + 2CO₂↑

(b) Action of sulphuric acid on sulphur

S + 2H₂SO₄→ 3SO₂+ 2H₂O

Solution C 14

ZnO + H₂SO₄→ ZnSO₄ + H₂O.

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Class 10 SELINA Solutions Chemistry Chapter 11 - Study of Compounds D. Sulphuric Acid

Study of Compounds D. Sulphuric Acid Exercise Intext 1

Solution 1

Solution 2

Solution 3

Solution 4

Study of Compounds D. Sulphuric Acid Exercise Ex. 11

Solution B 1

Solution B 2(a)

Solution B 2(b)

Solution B 2(e)

Solution B 3

Solution B 6

Solution B 9

Solution C 1

Solution C 2

Solution C 3

Solution C 5

Solution C 6

Solution C 9(c)

Solution C 11

Solution C 13

Solution D 1

Solution D 2(a)

Solution D 2(b)

Solution 2

Solution 3

Solution 4

Solution A 1

Solution A 2

Solution A 3

Solution B 2(c)

Solution B 2(d)

Solution B 2(f)

Solution B 2(g)

Solution B 2(h)

Solution B 2(i)

Solution B 4

Solution B 5

Solution B 7

Solution B 8

Solution C 4(a)

Solution C 4(b)

Solution C 10

Solution C 12

Solution C 14

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