Question

When 22.4 litres of $H_2\left(g\right)$ is mixed with 11.2 litres of $C{l}_2\left(g\right)$, each at $S.T.P,$ the moles of $HCl\left(g\right)$ formed is equal to :-

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Answer 1

$n_{H_2}=\frac{V\left(L\right)}{22.4}=\frac{22.4}{22.4}=1\text{mol}$

$n_{C{l}_2}=\frac{11.2}{22.4}=0.5\text{mol}$

$H_{2\left(g\right)}+C{l}_{2\left(g\right)}\to 2HC{l}_{\left(g\right)}$

initially - $1\text{mol}0.5\text{mol}0$

Here, $C{l}_2$ is the limiting reagent. Thus $C{l}_2$ will decide the moles of $HCl$ formed.
$\therefore$
Moles of $HCl$ formed $=\frac{2}{1}\times 0.5=1\text{mol}$

Hence, the correct answer is option (a).