Question

A water gas mixture has the composition by volume of $50 % H_{2}, 40 % CO$ and $5 % {CO}_{2} .$
(i) Calculate the volume in litres at STP of the mixture which on treatment with excess steam will contain $5$ litres of $H_{2} .$ The stoichiometry for the water gas shift reaction is

$CO + H_{2} O \to {CO}_{2} + H_{2}$

(ii) Find the density of the water gas mixture in $kg / m^{3} .$

(iii) Calculate the moles of the absorbants $KOH, Ca (OH)_{2}$ and ethanolamine.

$HO - {CH}_{2} - {CH}_{2} - {NH}_{2}$ required respectively to collect the ${CO}_{2}$ gas obtained.

Accepted Answer

(i) Let initial volume of mixture is 'V' then

$H_{2} \to 0.5 V, CO \to 0.45 V, {CO}_{2} \to 0.05 V$

On reaction with excess steam

$\begin{array}{l} CO + H_{2} O \to & {CO}_{2} + & H_{2} \\ 0.45 V & 0.05 V & 0.5 V \\ 0 & (0.05 + 0.45) V & (0.5 + 0.45) V \end{array}$

The volume of H₂ is 5 L

$0.95 V = 5$

$V = 5.263 L$

(ii) Molecular mass of initial mixture

$M = 0.5 \times 2 + 0.45 \times 28 + 0.05 \times 44 = 15.8$

$PV = nRT PV = \frac{m}{M} RT PM = ρRT$

$1 \times 15.8 = ρ \times 0.0821 \times 273 ρ = 0.7 g / L$

Hence. the density of the gas mixture is 0.7 g/L

(iii) Volume of ${CO}_{2} = 0.5 V = 0.5 5.263 = 2.6315 L$

moles of ${CO}_{2} = \frac{2.6315}{22.4} = 0.1174$

moles of $KOH$ required $= 0.1174 \times 2 = 0.2349$

moles of $Ca (OH)_{2}$ required $= 0.1174$

moles of $OH - {CH}_{2} - {CH}_{2} - {NH}_{2}$ required

$= 0.1174 \times 2 = 0.2349$

Important Questions on States of Matter: Gases and Liquids

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