$3.0 \mathrm{~g}$ of magnesium is burnt completely with $1.6 \mathrm{~g}$ of $\mathrm{O}_2$ in a clo

Question

$3.0 \mathrm{~g}$ of magnesium is burnt completely with $1.6 \mathrm{~g}$ of $\mathrm{O}_2$ in a closed vessel. Which reactant is left in excess and what is its exact mass? (Atomic weights: $\mathrm{Mg}=24, \mathrm{O}=16$)

Professor PC Thomas Classes · Answer

(c) : The balanced equation is $2\mathrm{Mg} + \mathrm{O}_2 ightarrow 2\mathrm{MgO}$. Moles of $\mathrm{Mg} = \frac{3.0}{24} = 0.125 \mathrm{~mol}$. Moles of $\mathrm{O}_2 = \frac{1.6}{32} = 0.05 \mathrm{~mol}$. According to the equation, 1 mole of $\mathrm{O}_2$ reacts with 2 moles of $\mathrm{Mg}$. Thus, $0.05 \mathrm{~mol}$ of $\mathrm{O}_2$ requires $2 	imes 0.05 = 0.10 \mathrm{~mol}$ of $\mathrm{Mg}$. Since $0.125 \mathrm{~mol}$ of $\mathrm{Mg}$ is available, $\mathrm{O}_2$ is the limiting reagent and $\mathrm{Mg}$ is in excess. The unreacted excess moles of $\mathrm{Mg} = 0.125 - 0.10 = 0.025 \mathrm{~mol}$. Mass of excess $\mathrm{Mg} = 0.025 	imes 24 = 0.6 \mathrm{~g}$.