According to reaction, $\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow…
According to reaction,
$\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_{2(\mathrm{aq})}+\mathrm{H}_{2(\mathrm{~g})} \uparrow$
Calculate the mass of $\mathrm{Mg}$ required to liberate
$4.48 \mathrm{dm}^3 \mathrm{H}_2$ at STP.
$\left(\right.$ Molar mass of $\mathrm{Mg}=24 \mathrm{~g} \mathrm{~mol}^{-1}$ )
$12 \mathrm{~g}$
$4.8 \mathrm{~g}$
$6 \mathrm{~g}$
$2.4 \mathrm{~g}$
Solution
Number of moles of a gas (n)
$=\frac{\text { Volume of gas at STP }}{22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}}$
$\therefore \quad \mathrm{n}=\frac{4.48 \mathrm{dm}^3}{22.4 \mathrm{dm}^3 \mathrm{~mol}^{-1}}=0.2 \mathrm{~mol}$
$\mathrm{Mg}_{(\mathrm{s})}+2 \mathrm{HCl}_{(\mathrm{aq})} \longrightarrow \mathrm{MgCl}_2+\mathrm{H}_{2(\mathrm{~g})} \uparrow$
$1 \mathrm{~mol} \mathrm{Mg} \equiv 1 \mathrm{~mol} \mathrm{H}_2$ gas
$\therefore \quad$ Mg required to liberate $0.2 \mathrm{~mol} \mathrm{H}_2$ gas
$=0.2 \mathrm{~mol}=0.2 \times 24=4.8 \mathrm{~g}$