Question

$F{e}_2{O}_3\left(s\right)$ may be converted to $Fe$ by the reaction
$F{e}_2{O}_3\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2Fe\left(s\right)+3H_{2}O\left(g\right)$
for which $K_c=8$ at temperature ${720}^{o}C$.
What percentage of the $H_2$ remains unreacted after the reaction has come to the equilibrium?

• A. $22\%$
• B. $34\%$
• C. $66\%$
• D. $78\%$

Answer 1

The correct option is B $34\%$

Let, initial moles of $H_2\left(g\right)$ is 1
$F{e}_2{O}_3\left(s\right)+3H_2\left(g\right)\rightleftharpoons 2Fe\left(s\right)+3H_{2}O\left(g\right)|atequilibrium-1-3x3x$
$K_c=\frac{(\frac{3x}{V}{)}^3}{(\frac{1-3x}{V}{)}^3}$
$\Rightarrow 8=(\frac{3x}{1-3x}{)}^3$
$\Rightarrow x=0.22$
$\%of{H}_2\text{unreacted}=\frac{1-3\times 0.22}{1}\times 100=34\%$