Chemistry - Stoichiometry Question with Solution | TestHub

$\mathrm{Mg}+\frac{1}{2}{\mathrm{O}}_2⟶\mathrm{MgO}$

Moles of $\mathrm{Mg}=\frac{12}{24}=\frac{1}{2}$

Mass of ${\mathrm{O}}_2=\frac{1}{4}\times 32=8\mathrm{g}$

Mass of air $=8+1\times 28=36\mathrm{g}$

$3\mathrm{Mg}+{\mathrm{N}}_2⟶{\mathrm{Mg}}_3{\mathrm{N}}_2$

Mass of nitrogen $=\frac{1}{6}\times 28=\frac{28}{6}\mathrm{g}$

Mass of air $=\frac{28}{6}+\frac{1}{24}\times 32=6\mathrm{g}$

$\mathrm{Mg}+\frac{1}{2}{\mathrm{O}}_2⟶\mathrm{MgO}$

$3\mathrm{Mg}+{\mathrm{N}}_2⟶{\mathrm{Mg}}_3{\mathrm{N}}_2$

If ' x ' moles of Mg reacts for MgO , then ( $0.5-\mathrm{x}$ ) moles of Mg reacts for ${\mathrm{Mg}}_3{\mathrm{N}}_2$. $\frac{1}{3}\left(\frac{1}{2}-x\right)=4\times \frac{x}{2}\Rightarrow x=\frac{1}{14}$ Mass of air needed $=\frac{1}{28}\times 32+\frac{1}{3}\left(\frac{1}{2}-\frac{1}{14}\right)\times 28=5.14\mathrm{g}$ Total mass of products $=12=5.14=17.14\mathrm{g}$.