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 STP, the moles of HCl(g) formed is equal to :

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

Answer 1

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

At STP, $1$ mole of any gas occupies a volume of $22.4$ L.

$\underset{22.4lt}{H_2}+\underset{11.2lt}{C{l}_2}\to 2HCl$

Limiting reagent is $C{l}_2$.
$1$ mole of chlorine forms $2$ moles of HCl.

Hence, $0.5$ moles of chlorine will form $1$ mole of HCl.

Thus, when $22.4$ liters of $H_2\left(g\right)$ is mixed with $11.2$ liters of $C{l}_2\left(g\right)$, each at STP, the moles of HCl(g) formed is equal to $1$ mol of HCl(g).