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(g) formed is equal to:

• A. 0.5 mol of HCl (g)
• B. 1.5 mol of HCl (g)
• C. 1 mol of HCl (g)
• D. 2 mol of HCl (g)

Answer 1

The correct option is C 1 mol of HCl (g)

1 mole = 22.4 liters at S.T.P.

${}^n{H}_2=\frac{22.4}{22.4}=1mol;{}^{n}C{l}_2=\frac{11.2}{22.4}=0.5mol$

Reaction is as,

$\begin{array}{cccccc}& H_{2\left(g\right)}& +& C{l}_{2\left(g\right)}& ⟶& 2HC{l}_{\left(g\right)}\\ \text{Initial}& 1mol& & 0.5mol& & 0\\ \text{Final}& (1-0.5)& & (0.5-0.5)& & 2\times 0.5\\ & =0.5mol& & =0mol& & 1mol\end{array}$

Here, $C{l}_2$ is limiting reagent. So, 1 mole of HCl(g) is formed.