Stoichiometry

Q: 10 ml of a solution of H₂O₂ labelled ’10 volume’ just decolorises 100 ml of potassium permanganate solution acidified with dilute H₂SO₄. Calculate the amount of potassium permanganate in the given solution.

(A) 0.1563 gm

(B) 0.563 gm

(C) 5.63 gm

(D) 0.256 gm

Sol:

2H₂O₂ → H­₂O + O₂

22400 ml of O₂ evolved from 68 gm of H₂O₂

∴ 10 ml of O_­2 is evolved from gm of H₂O₂

Hence 1 ml  of H₂O₂ contain  mol = 0.00178 equivalent 10 ml of H₂O₂ will 0.0178 equivalents which will be  present in 100 ml of KMnO₄ solution.

Amount of KMnO­₄ in given sample =  = 0.563 gm

            Ans:  (B)

Q: The hourly energy requirement of an astronaut can be satisfied by the energy released when 34 g of sucrose are burnt in his body. How many g of oxygen would be needed to be carried in space capsule to meet his requirement for one day?

(A) 916.2 gm

(B) 91.62 gm

(C) 8.162 gm

(D) 9.162 gm

Sol: C₁₂H₂₂O₁₁ + 12O₂ ¾® 12CO₂ + 11H₂O

1 mole of starch requires 12 mole of oxygen

mole of starch requires  = 916.2 gm

Ans: (A)

Q: Calculate the weight of ion which will be converted into its oxide by the action of 18 g of steam on it.

(A) 37.3 gm

(B) 3.73 gm

(C) 56 gm

(D) 5.6 gm

Sol: 2Fe + 3H₂O → Fe₂O₃

3 × 18gm of steam will react with 2 × 56 gm of Iron

∴ 18 gm of steam will convert =$\frac{18 \times 2 \times 56}{3 \times 18}$

= 37.3 gm of Fe

Ans: (A)

Q: 25.4 g of iodine and 14.2g of chlorine are made to react completely to yield a mixture of ICl and ICl₃. Calcualte the ratio of moles of ICl and ICl₃.

(A) 1:1

(B) 1:2

(C) 1:3

(D) 2:3

Sol: I₂ + 2Cl₂ → ICl + ICl₃

25.4 gm of Iodine contains 0.1 moles of it

14.2 gm of chlorine contains 0.2 moles of it

∴ Moles of ICl & ICl₃ produced will be 0.1 and 0.1. Hence Molar ratio 1: 1

Ans: (A)

Q: The volume in ml of 0.1 N HCl required to react completely with 1.0 gm mixture of Na₂CO_­3 and NaHCO_­3 containing equimolar amounts of the two compounds is

(A) 157.9 ml

(B) 152.6 ml

(C) 200 ml

(D) 98.5 ml

Sol: Let amount of Na₂CO₃ in 1 gm of mixture be a and since Na₂CO₃ and NaHCO₃ are in equimolar amounts.

$\frac{a}{106} = \frac{1-a}{84} $

a = 0.588 gm

∴ Weight of NaHCO₃ = (1 – 0.558) = 0.442 gm

Met of Na₂CO₃ + Meq. of NaHCO₃ = Meq. of HCl

$\large \frac{0.588}{53} \times 1000 + \frac{0.494}{84} \times 100 = 0.1 \times V$

V = 157.9 ml

 Ans: (A)