Question

$60\text{g}$ of water gas and $18\text{g}$ of steam are heated in a closed vessel to a temperature of ${450}^{o}C$ so that the equilibrium:
$C{O}_{\left(g\right)}+H_2{O}_{\left(g\right)}\rightleftharpoons C{O}_{2\left(g\right)}+H_{2\left(g\right)}$ is reached. If $K$ for the reaction is $3$, the mass of $C{O}_2$ present will be:

• A. $33\text{g}$
• B. $22\text{g}$
• C. $44\text{g}$
• D. $60\text{g}$

Answer 1

The correct option is A $33\text{g}$

Mass of water gas $=60g(CO+H_{2}O)$
Mass of steam $\left(H_{2}O\right)=18g$
Mass of $CO=60-18=42g$Moles of $CO=\frac{42}{28}=1.5moles$
Moles of $H_{2}O=\frac{18}{18}=1mole$
$C{O}_{\left(g\right)}+H_2{O}_{\left(g\right)}\rightleftharpoons C{O}_{2\left(g\right)}+H_{2\left(g\right)}$
$1.5$ $1$ $-$ $-$
Equilibrium $1.5-x$ $1-x$ $x$ $x$
$K=\frac{x^2}{(1.5-x)(1-x)}=3$
$x^2=3(x^2-2.5x+1.5)$
$x^2=3x^2-7.5x+4.5$
$2x^2-7.5x+4.5=0$
$x=3;$ [Rejected] $x=0.75$
$\therefore \frac{3}{4}moles$ of $C{O}_2$ are present.
Mass $=\frac{3}{4}\times 44=33g$