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 $\frac{680}{22400}$ gm of H₂O₂

Hence 1 ml  of H₂O₂ contain  $ \frac{0.0303}{34} 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 $=\frac{158}{5} \times 0.00178 $  = 0.563 gm

Ans: (B)

Q: The hourly energy requirement of an astronaut can be satisfied by the energy released when 34g 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  → 12CO₂ + 11H₂O

1 mole of starch requires 12 mole of oxygen

∴ $\large \frac{34}{342}$ mole of starch requires $\large \frac{34}{342} \times 12 \times 32 \times 24 $

= 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  $\large = \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.

$\large \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

$\large \frac{0.588}{53} \times 1000 + \frac{0.492}{84}\times 100 = 0.1 \times V$

∴ V = 157.9 ml

Ans:  (A)

Q: A mixture of KBr and NaBr weighing 0.560 gm was treated with aqueous Ag⁺ and all the bromide ion was recovered as 0.970 gm of pure AgBr. The fraction by weight of KBr in the sample is

(A) 0.25 gm

(B) 0.2378

(C) 0.36

(D) 0.285

Sol:  KBr + NaBr          +  Ag⁺  →AgBr

a gm    (0.56 – a)                    0.97 gm

Applying POAc for Br atom 1 × n_KBr  + 1 × n_NaBr = 1 × n_AgBr

$\large \frac{a}{119} + \frac{0.560 a}{103} = \frac{0.97}{188} $

∴  a = 0.1332 gm

∴ Fraction of KBr in the sample = 0.1332/0.560 = 0.2378

Ans: (B)

Q: 25 ml of H₂O₂ solution were added to excess of acidified solution of KI and iodine so liberated required 20 ml of 0.1 N Na₂S₂O₃ for titration. The normality H₂O₂ is ?

(A) 0.02

(B) 0.04

(C) 0.08

(D) 0.03

Sol:

O₂^–1 + 2e → 2O^–2                   I­₂ + S₂O₃^2–  → 2I^– + S₄O₆^2–

2I^–  → I₂ + 2e

Meq. of H₂O₂  = Meq. of I₂ = Meq of Na₂S₂O₃

N × 25 = 0.1 × 20

$N_{H_2 O_2} = 0.08 $

Ans: (C)