Mole Concept And Stoichiometry Exe-5(A) Concise Class-10 ICSE Chemistry Selina Solutions Chapter-5. We Provide Step by Step Answer of Exe-5(A) 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(A) 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	Mole Concept And Stoichiometry
Topics	Exe-5(A)
Edition	2025-2026

Exe-5(A) Questions on Mole Concept And Stoichiometry

Page-77

Que-1: 

(a) Calculate the volume of oxygen at S.T.P required for the complete combustion of 100 litres of carbon monoxide at the same temperature and pressure.
2CO + O₂    -> 2CO₂
(b) 200 cm³ of hydrogen and 150 cm³ of oxygen are mixed and ignited, as per the following reaction,
2H₂ + O₂  -> 2H₂O
What volume of oxygen remains unreacted?

Ans:
(a) 2CO + O₂  -> 2CO₂
2 V    1 V     2 V
2 V of CO requires = 1V of O₂
so, 100 litres of CO requires = 50 litres of O₂

(b) 2H₂ + O₂   -> 2H₂O
2 V         1V            2V
From the equation, 2V of hydrogen reacts with 1V of oxygen
so 200cm³ of Hydrogen reacts with = 200/2= 100 cm³
Hence, the unreacted oxygen is 150 – 100 = 50cm³ of oxygen.

Que-2: 24 cc Marsh gas (CH₄) was mixed with 106 cc oxygen and then exploded. On cooling the volume of the mixture became 82 cc, of which, 58 cc was unchanged oxygen. Which law does this experiment support? Explain with calculation.

Ans:
This experiment supports Gay lussac’s law of combining volumes.
Since the unchanged or remaining O₂ is 58 cc so, used oxygen 106 – 58 = 48cc
According to Gay lussac’s law, the volumes of gases reacting should be in a simple ratio.
CH₄+ 2O₂    -> CO₂ + 2H₂O
1 V2 V
24 cc48 cc
i.e. methane and oxygen react in a 1:2 ratio.

Que-3: What volume of oxygen would be required to burn completely 400 ml of acetylene [C₂H₂]? Also calculate the volume of carbon dioxide formed.

2C₂H₂ + 5O₂   

4CO₂ + 2H₂O (l)

Ans:
2C₂H₂ + 5O₂    4CO₂ + 2H₂O (l)
2 V        5 V           4 V
From equation, 2 V of C₂H₂ requires = 5 V of O₂
So, for 400ml C₂H₂ , O₂ required = 400  5/2 =1000 ml
Similarly, 2 V of C₂H₂ gives = 4 V of CO₂
So, 400ml of C₂H₂ gives CO₂ = 400   4/2 = 800ml

Que-4: 112 cm³ of H₂S(g) is mixed with 120 cm³ of Cl₂(g) at STP to produce HCl(g) and sulphur(s). Write a balanced equation for this reaction and calculate (i) the volume of gaseous product formed (ii) composition of the resulting mixture.

Ans:
Balanced chemical equation:
H2S + Cl2 -> 2HCl + S
(i) At STP, 1 mole gas occupies 22.4 L.
As 1 mole H₂S gas produces 2 moles HCl gas,
22.4 L H₂S gas produces 22.4 × 2 = 44.8 L HCl gas.
Hence, 112 cm³ H₂S gas will produce 112 × 2 = 224 cm³ HCl gas.
(ii) 1 mole H₂S gas consumes 1 mole Cl₂ gas.
This means 22.4 L H₂S gas consumes 22.4 L Cl₂ gas at STP.
Hence, 112 cm³ H₂S gas consumes 112 cm³ Cl₂ gas.
120 cm³ – 112 cm³ = 8 cm³ Cl₂ gas remains unreacted.
Thus, the composition of the resulting mixture is 224 cm³HCl gas + 8 cm³ Cl₂ gas.

Que-5:  1250cc of oxygen was burnt with 300cc of ethane [C₂H₆]. Calculate the volume of unused oxygen and the volume of carbon dioxide formed; 2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O

Ans:
2C₂H₆ + 7O₂ ⟶ 4CO₂ + 6H₂O
From the equation, 2V of ethane reacts with 7V oxygen.
So, 300 cc of ethane reacts with 300 x 7/2 = 1050 cc
Hence, unused O₂ = 1250 – 1050 = 200 cc
From 2V of ethane, 4V of CO₂ is produced.
So, 300 cc of ethane will produce 300 x 4/2 = 600 cc of CO2

Que-6: What volume of oxygen at STP is required to affect the combustion of 11 litres of ethylene [C₂H₄] at 273^o C and 380 mm of Hg pressure?

C₂H₄+3O₂   -> 2CO₂ + 2H₂O

Ans:
C₂H₄+3O₂  -> 2CO₂ + 2H₂O

STP	Given Values
P1 = 760 mm of Hg	P2 = 380 mm of Hg
V1 = x lit	V2 = 33 lit
T1 = 273 K	T2 = 273 + 273 K = 546 K

Using the gas equation,

Therefore, volume of oxygen required = 8.25 lit.

Que-7: Calculate the volume of HCl gas formed and chlorine gas required when 40 ml of methane reacts completely with chlorine at S.T.P.

CH₄ + 2Cl₂  -> CH₂Cl₂+2HCl

Ans:
CH₄ + 2Cl₂  ->  CH₂Cl₂ +2HCl
1 V      2 V            1 V       2 V
From equation, 1V of CH₄ gives = 2 V HCl
so, 40 ml of methane gives = 80 ml HCl
For 1V of methane = 2V of Cl₂ required
So, for 40ml of methane = 40  2 = 80 ml of Cl₂

Que-8: What volume of propane is burnt for every 500 cm³ of air used in the reaction under the same conditions? (assuming oxygen is 1/5^th of air)

C₃H₈ + 5O₂   -> 3CO₂ + 4H₂O

Ans:
C₃H₈ + 5O₂  -> 3CO₂ + 4H₂O
1 V        5 V        3 V
From equation, 5 V of O₂ required = 1V of propane
_{so, 100 cm³ of O2 will require = 20 cm³ of propane}

Que-9: 450 cm³ of nitrogen monoxide and 200 cm³ of oxygen are mixed together and ignited. Caclulate the composition of resulting mixture.

2NO + O₂ -> 2NO₂

Ans:
2NO + O₂    2NO₂
2 V       1 V         2 V
From equation, 1V of O₂ reacts with = 2 V of NO
200cm³ oxygen will react with = 200 2 =400 cm³ NO
Hence, remaining NO is 450 – 400 = 50 cm³
NO₂ produced = 400cm³ because 1V oxygen gives 2 V NO₂
Total mixture = 400 + 50 = 450 cm³

Que-10: If 6 liters of hydrogen and 4 liters of chlorine are mixed and exploded and if water is added to the gases formed, find the volume of the residual gas.

Ans:
(i) 6 litres of hydrogen and 4 litres of chlorine when mixed, results in the formation of 8 litres of HCl gas.
(ii) When water is added to it, it results in the formation of hydrochloric acid. Chlorine acts as a limiting agent leving behind only 2 litres of hydrogen gas.
(iii) Therefore, the volume of the residual gas will be 2 litres.

Que-11: Ammonia may be oxidised to nitrogen monoxide in the presence of a catalyst according to the following equation.

4NH₃ + 5O₂   -> 4NO + 6H₂O
If 27 litres of reactants are consumed, what volume of nitrogen monoxide is produced at the same temperature and pressure?

Ans:
4NH₃ + 5O₂  -> 4NO + 6H₂O
4 V        5 V         4 V
9 litres of reactants gives 4 litres of NO
So, 27 litres of reactants will give = 27 x  4/9 = 12 litres of NO

Que-12: A mixture of hydrogen and chlorine occupying 36 cm³ was exploded. On shaking it with water, 4cm³ of hydrogen was left behind. Find the composition of the mixture.

Ans:
H₂ + Cl₂    -> 2HCl
1V    1V            2 V
Since 1 V hydrogen requires 1 V of oxygen and 4cm³ of H₂ remained behind so the mixture had composition: 16 cm³ hydrogen and 16 cm³ chlorine.
Therefore Resulting mixture is H₂ =4cm³,HCl=32cm³

Que-13: What volume of air (containing 20% O₂ by volume) will be required to burn completely 10 cm³ each of methane and acetylene?

CH₄ + 2O₂   ->CO₂ + 2H₂O
2C₂H₂ + 5O₂ -> 4CO₂ + 2H₂O

Ans:
CH₄ + 2O₂  -> CO₂ + 2H₂O
1 V       2 V          1 V
2C₂H₂ + 5O₂    ->  4CO₂ + 2H₂O
2 V          5 V            4 V
From the equations, we can see that
1V CH₄ requires oxygen = 2 V O₂
So, 10cm³ CH₄ will require =20 cm³ O₂
Similarly 2 V C₂H₂ requires = 5 V O₂
So, 10 cm³ C₂H₂ will require = 25 cm³ O₂
Now, 20 V O₂ will be present in 100 V air and 25 V O₂ will be present in 125 V air ,so the volume of air required is 225cm³

Que-14: LPG has 60% propane and 40% butane: 10 litres of this mixture is burnt. Calculate the volume of carbon dioxide added to atmosphere.

C₃H₈ + 5O₂  -> 3CO₂ + 4H₂O
2C₄H₁₀ + 13O₂  ->  8CO₂ + 10H₂O

Ans:
C₃H₈ + 5O₂   -> 3CO₂ + 4H₂O
2C₄H₁₀ + 13O₂  -> 8CO₂ + 10H₂O
60 ml of propane (C₃H₈) gives 3 x 60 = 180 ml CO₂
40 ml of butane (C₄H₁₀) gives = 8 x  40/2 = 160 ml of CO₂
Total carbon dioxide produced = 340 ml
So, when 10 litres of the mixture is burnt = 34 litres of CO₂ is produced.

Que-15: Water decomposes to O₂ and H₂ under suitable conditions as represented by the equation below:

2H₂O ⟶ 2H₂ + O₂
(a) If 2500 cm³ of H₂ is produced, what volume of O₂ is liberated at the same time and under the same conditions of temperature and pressure?
(b) The 2500 cm³ of H₂ is subjected to 2(1/2) times increase in pressure (temp. remaining constant). What volume will H₂ now occupy?
(c) Taking the value of H₂ calculated in 5(b), what changes must be made in Kelvin (absolute) temperature to return the volume to 2500 cm³ pressure remaining constant.

Ans: 
2H₂O ⟶ 2H₂ + O₂
2 Vol. of water gives 2 Vol. of H₂ and 1 Vol. of O₂
∴ If 2500 cm³ of H₂ is produced, volume of O₂ produced = 2500/2 = 1250 cm³
(b) V₁ = 2500 cm³
P₁ = 1 atm = 760 mm
T₁ = T
T₂ = T
P₂ = [760 x 2(1/2)] + [760] = 760 [5/2 + 1] = 760 x 7/2 = 2660 mm
V₂ = ?
Using formula:

P₁ = P₂ = P
T₁ = T
V₂ = 2500 cm³
T₂ = ?
Using formula:

T₂ = 3.5 x T
Therefore , T₂ = 3.5 times T or temperature should be increased by 3.5 times

Que-16: The gases chlorine, nitrogen, ammonia and sulphur dioxide are collected under the same conditions of temperature and pressure. The following table gives the volumes of gases collected and the number of molecules (x) in 20 litres of nitrogen. You are to complete the table giving the number of molecules in the other gases in terms of x.

Gas	Volume (in litres)	Number of molecules
Chlorine Nitrogen Ammonia Sulphur dioxide	10 20 5	x

Ans:

Gas	Volume (in litres)	Number of molecules
Chlorine	10	x/2
Nitrogen	20	x
Ammonia	20	X
Sulphur dioxide	5	x/4

 

Que-17: 

(i) If 150 cc of gas A contains X molecules, how many molecules of gas B will be present in 75 cc of B?
The gases A and B are under the same conditions of temperature and pressure.
(ii) Name the law on which the above problem is based.

Ans:
(i) According to Avogadro’s law, under the same conditions of temperature and pressure, equal volumes of different gases have the same number of molecules.
As 150 cc of gas A contains X molecules, 150 cc of gas B also contains X molecules.
So, 75 cc of B will contain X/2 molecules.
(ii) The problem is based on Avogadro’s law.

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