Mole Concept And Stoichiometry Exe-5(D) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5. We Provide Step by Step Answer of Exe-5(D) Questions of Exercise-5 for ICSE Class-10. The given Solutions is according to the Latest editions. Visit official Website CISCE for detail information about ICSE Board Class-10.

Mole Concept And Stoichiometry Exe-5(D) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5

Board	ICSE
Book / Publication	Concise / Selina
Subject	Chemistry
Class	10th
Writer	Dr SP Singh
Chapter-5	Acids Bases and Salts
Topics	Exe-5(D)
Edition	2025-2026

Exe-5(D) Questions on Stoichiometry

Page-94

Que-1: The reaction between 15 g of marble and nitric acid is given by the following equation:

CaCO₃ + 2HNO₃  Ca(NO₃)₂+ H₂O + CO₂
Calculate: (a) the mass of anhydrous calcium nitrate formed, (b) the volume of carbon dioxide evolved at S.T.P.

Ans:
(a) 100 g of CaCO₃ produces = 164 g of Ca(NO₃)₂
So, 15 g CaCO₃ will produce = 164 15/100 = 24.6 g Ca(NO₃)₂
(b) 1 V of CaCO₃ produces 1 V of CO₂
100 g of CaCO₃ has volume = 22.4 litres
So, 15 g will have volume = 22.4 15/100 = 3.36 litres CO₂

Que-2: 66g of ammonium sulphate is produced by the action of ammonia on sulphuric acid.

Write a balanced equation and calculate:
(a) Mass of ammonia required.
(b) The volume of the gas used at the S.T.P.
(c) The mass of acid required.

Ans:
2NH₃ + H₂SO₄ (NH₄)₂SO₄
66 g
(a) 2NH₃ + H₂SO₄ (NH₄)₂SO₄
34 g98 g132 g
For 132 g (NH₄)₂SO₄ = 34 g of NH₃ is required
So, for 66 g (NH₄)₂SO₄ = 66 32/132 = 17 g of NH₃ is required
(b) 17g of NH₃ requires volume = 22.4 litres
(c) Mass of acid required, for producing 132g (NH₄)₂SO₄ = 98g
So, Mass of acid required, for 66g (NH₄)₂SO₄ = 66 98/132 = 49g

Que-3: The reaction between red lead and hydrochloric acid is given below:

Pb₃O₄ + 8HCl  3PbCl₂ + 4H₂O + Cl₂
Calculate:
(a) the mass of lead chloride formed by the action of the 6.85 g of red lead,
(b) the mass of the chlorine and
(c) the volume of the chlorine evolved at S.T.P.

Ans:
(a) Molecular mass of Pb₃O₄ = 3 207.2 + 4 16 = 685 g
685 g of Pb₃O₄ gives = 834 g of PbCl₂
Hence, 6.85 g of Pb₃O₄ will give = 6.85 834/685 = 8.34 g
(b) 685g of Pb₃O₄ gives = 71g of Cl₂
Hence, 6.85 g of Pb₃O₄ will give = 6.85 71/685 = 0.71 g Cl₂
(c) 1 V Pb₃O₄ produces 1 V Cl₂
685g of Pb₃O₄has volume = 22.4 litres = volume of Cl₂ produced
So, 6.85 Pb₃O₄ will produce = 6.85 22.4/685 = 0.224 litres of Cl₂

Que-4: Find the mass of KNO₃ required to produce 126 kg of nitric acid. Find whether a larger or smaller mass of NaNO₃ is required for the same purpose.

KNO₃+ H₂SO₄    KHSO₄ + HNO₃
NaNO₃ + H₂SO₄   NaHSO₄ + HNO₃

Ans:
Molecular mass of KNO₃ = 101 g
63 g of HNO₃ is formed by = 101 g of KNO₃
So, 126000 g of HNO₃ is formed by = 126000 101/63 = 202 kg
Similarly,126 g of HNO₃ is formed by 170 kg of NaNO₃
So, smaller mass of NaNO₃ is required.

Que-5: Pure calcium carbonate and dilute hydrochloric acid are reacted and 2 litres of carbon dioxide was collected at 27^oC and normal pressure.

Calculate :
(a) The mass of salt required.
(b) The mass of the acid required to prepare the 2 litres of CO₂ at 27 C and normal pressure.
CaCO₃ + 2HCl   CaCl₂ + H₂O + CO₂

Ans:
CaCO₃ + 2HCl   CaCl₂ + H₂O + CO₂
100g73g22.4L
(a) V₁ =2 litresV₂ =?
T₁ = (273+27)=300KT₂=273K
V₁/T₁=V₂/T₂
V₂=V₁T₂/T₁= [(2 x 273)/300]L
Now at STP 22.4 litres of CO₂ are produced using CaCO₃ =100g
So, [(2 x 273)/300]    litres are produced by =100/22.4 2274/300 =.125g
(b)  22.4 litres are CO₂ are prepared from acid =73g
[(2 x 273)/300] litres are prepared from = 73/22.4 2273/300=5.9g

Que-6: Calculate the mass and volume of oxygen at S.T.P., which will be evolved on electrolysis of 1 mole (18g) of water.

Ans: 
2H₂O  –> 2H₂ + O₂
2 V2 V1 V
2 moles of H₂O gives = 1 mole of O₂
So, 1 mole of H₂O will give = 0.5 moles of O₂
so, mass of O₂ = no. of moles x molecular mass
= 0.5 32 = 16 g of O₂
and 1 mole of O₂ occupies volume =22.4 litre
so, 0.5 moles will occupy = 22.4   0.5 = 11.2 litres at S.T.P.

Que-7: 1.56 g of sodium peroxide reacts with water according to the following equation:

2Na₂O₂ + 2H₂O     4NaOH + O₂
Calculate:
(a) mass of sodium hydroxide formed,
(b) Volume of oxygen liberated at S.T.P.
(c) Mass of oxygen liberated.

Ans:
2Na₂O₂ + 2H₂O   4NaOH + O₂
2 V4 V1 V
(a) Mol. Mass of Na₂O₂ = 2 23 + 2 16 = 78 g
Mass of 2Na₂O₂= 156 g
156 g Na₂O₂ gives = 160 g of NaOH (4 40 g)
So, 1.56 Na₂O₂ will give = 160 1.56/156 = 1.6 g
(b) 156 g Na₂O₂ gives = 22.4 litres of oxygen
So, 1.56 g will give = 22.4 1.56/156 = 0.224 litres
= 224 cm³
(c)156 g Na₂O₂ gives = 32 g O₂
So, 1.56 g Na₂O₂ will give = 32 1.56/156
= 32/100 = 0.32 g

Que-8: 

(a) Calculate the mass of ammonia that can be obtained from 21.4 g of NH₄Cl by the reaction:
2NH₄Cl + Ca(OH)₂     CaCl₂ +2H₂O + 2NH₃
(b) What will be the volume of ammonia when measured at S.T.P?
The molar volume of a gas = 22.4 litres at STP.

Ans:
2NH₄Cl + Ca(OH)₂  CaCl₂+2H₂O + 2NH₃
2 V1 V1 V2 V
Mol. Mass of 2NH₄Cl = 2[14 + (1 4) + 35.5] = 2[53.5] = 107 g
(a) 107 g NH₄Cl gives = 34 g NH₃
So, 21.4 g NH₄Cl will give = 21.4 34/107 = 6.8 g NH₃
(b) The volume of 17 g NH₃ is 22.4 litre
So, volume of 6.8 g will be = 6.8 22.4/17 = 8.96 litre

Que-9: Aluminium carbide reacts with water according to the following equation.

Al₄C₃ + 12H₂O → 4Al(OH)₃ + 3CH₄
(a) What mass of aluminium hydroxide is formed from 12g of aluminium carbide?
(b) What volume of methane s.t.p. is obtained from 12g of aluminium carbide? (2018)

Ans:

Que-10: MnO₂ + 4HCl   MnCl₂ + 2H₂O +Cl₂

0.02 moles of pure MnO₂is heated strongly with conc. HCl. Calculate:
(a) mass of MnO₂ used
(b) moles of salt formed,
(c) mass of salt formed,
(d) moles of chlorine gas formed,
(e) mass of chlorine gas formed,
(f) volume of chlorine gas formed at S.T.P.,
(g) moles of acid required,
(h) Mass of acid required.

Ans:
MnO₂ + 4HCl   MnCl₂ + 2H₂O +Cl₂
1 V4 V1 V1 V
(a) 1 mole of MnO₂ weighs = 87 g (mol. Mass)
So, 0.02 mole will weigh = 87 0.02 = 1.74 g MnO₂
(b) 1 mole MnO₂ gives = 1 mole of MnCl₂
So, 0.02 mole MnO₂will give =0.02 mole of MnCl₂
(c) 1 mole MnCl₂ weighs = 126 g(mol mass)
So, 0.02 mole MnCl₂ will weigh = 126 0.02 g = 2.52 g
(d) 0.02 mole MnO₂will form =0.02 mole of Cl₂
(e) 1 mole of Cl₂ weighs = 35.5 g
So, 0.02 mole will weigh = 71 0.02 = 1.42 g of Cl₂
(f) 1 mole of chlorine gas has volume = 22.4 litres
So, 0.02 mole will have volume = 22.4  0.02 = 0.448 litre
(g) 1 mole MnO₂requires HCl = 4 mole
So, 0.02 mole MnO₂ will require =4 0.02 = 0.08 mole
(h) For 1 mole MnO₂ , acid required = 4 mole of HCl
So, for 0.02 mole, acid required = 4 0.02 =0.08 mole
Mass of HCl = 0.08 x 36.5 = 2.92 g

Que-11: Nitrogen and hydrogen react to form ammonia.

N₂ (g) + 3H₂ (g)  2NH₃ (g)
If 1000g H₂ react with 2000g of N₂:
(a) Will any of the two reactants remain unreacted? If yes, which one and what will be its mass?
(b) Calculate the mass of ammonia(NH₃) that will be formed?

Ans:
N₂ + 3H₂   2NH₃
28g6g34g
28g of nitrogen requires hydrogen = 6g
2000g of nitrogen requires hydrogen = 6/28    2000=3000/7g
So mass of hydrogen left unreacted =1000-3000/7=571.4g of H₂
(b) 28g of nitrogen forms NH₃ = 34g
2000g of N₂ forms NH₃
= 34/28 2000
=2428.6g

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