Calculate the amount of $H_2$ required to completely react with 112 litres of $C{l}_2$ at STP to produce $HCl$:

The correct option is C $10\text{g}$
$H_2+C{l}_2\to 2HCl$
moles of $C{l}_2$ = mol of $H_2$
moles of $C{l}_2\text{taken}=\frac{112}{22.4}=5mol$
So, mol of $H_2\text{required}=5mol$
Amount of $H_2$ required = Moles $\times$ Molar mass
Mass of $H_2$ required = $5\text{mol}\times 2\text{g/mol}=10\text{g}$