Question

$3.0$ g of magnesium is burnt with $0.48\text{g}{O}_2$ in a closed vessel. Which reactant is left in excess and how much?
(Atomic mass of Mg and O are 24 u and 16 u respectively)

• A. $Mg,2.28\text{g}$
• B. $O_2,1.6\text{g}$
• C. $Mg,4.4\text{g}$
• D. $O_2,2.28\text{g}$

Answer 1

The correct option is A $Mg,2.28\text{g}$

$\text{Moles of Magnesium}\left(n_{Mg}\right)=\frac{3}{24}=0.125\text{mol}$
$\text{Moles of Oxygen molecule}\left(n_{O_2}\right)=\frac{0.48}{32}=0.015\text{mol}$
The balanced equation is
$\begin{array}{cccccc}& 2Mg& +& O_2& ⟶& 2MgO\\ \text{Initial}& 0.125\text{mole}& & 0.015\text{mole}& & 0\\ \text{Final}& (0.125-2\times 0.015)& & 0& & 2\times 0.015\end{array}$
Here, $O_2$ is limiting reagent.
$\therefore$ Mass of Mg left in excess $=0.095\times 24=2.28g$