$1.8 \mathrm{~g}$ of $\mathrm{Mg}$ is burnt in a closed vessel which contains $0.8 \mathrm{~g}$ of oxygen. Which of the following statements is not correct? Given ; $M(\mathrm{Mg})=24.3 \mathrm{~g} \mathrm{~mol}^{-1}$.

Question

Aicharya · Accepted Answer

The reaction is $2 \mathrm{Mg}+\mathrm{O}_{2} ightarrow 2 \mathrm{MgO}$
$\xi_{\max }(\mathrm{Mg})=\frac{m_{\mathrm{Mg}} / M_{\mathrm{Mg}}}{v_{\mathrm{Mg}}}=\frac{1.8 \mathrm{~g} / 24.3 \mathrm{~g} \mathrm{~mol}^{-1}}{2}=0.037 \mathrm{~mol}$
$\xi_{\max }\left(\mathrm{O}_{2}ight)=\frac{0.8 \mathrm{~g} / 32 \mathrm{~g} \mathrm{~mol}^{-1}}{1}=0.025 \mathrm{~mol}$
Since $\xi_{\max }\left(\mathrm{O}_{2}ight) < \xi_{\max }(\mathrm{Mg})$, the limiting reagent is oxygen.
$$
\Delta n_{\mathrm{MgO}}=v_{\mathrm{MgO}} \xi_{\max }\left(\mathrm{O}_{2}ight)=2(0.025 \mathrm{~mol})=0.05 \mathrm{~mol}
$$
Mass of $\mathrm{Mg}$ consumed $=\left(\frac{2 M_{\mathrm{Mg}}}{M_{\mathrm{O}_{2}}}ight) m_{\mathrm{O}_{2}}=\frac{48.6}{32} \times 0.8 \mathrm{~g}=1.22 \mathrm{~g}$
Mass of $\mathrm{Mg}$ left behind $=1.8 \mathrm{~g}-1.22 \mathrm{~g}=0.58 \mathrm{~g}$ *

$1.8 \mathrm{~g}$ of $\mathrm{Mg}$ is burnt in a closed vessel which contains $0.8 \mathrm{~g}$ of oxygen.…

Solution