Question

$1.0g$ of magnesium is burnt with $0.56g{O}_2$ in a closed vessel. Which reactant is left in excess and how much?
$(\text{At. wt. Mg}=24;O=16)$

• A. $\text{Mg},0.16g$
• B. $O_2,0.16g$
• C. $\text{Mg},0.44g$
• D. $O_2,0.28g$

Answer 1

The correct option is A $\text{Mg},0.16g$

$n_{Mg}=\frac{1}{24}=0.0416\text{mol}$

$n_{O_2}=\frac{0.56}{32}=0.0175\text{mol}$

$Mg\left(s\right)+\frac{1}{2}{O}_2\left(g\right)\to MgO\left(s\right)$

Initial: $0.04160.01750$
moles $\text{mol}\text{mol}$
$\therefore$ $O_2$ is the limiting reagent and $Mg$ is the excess reagent.
Thus, the moles of $Mg$ consumed $=0.0175\times 2=0.0350\text{mol}$
The moles of excess reagent $\left(Mg\right)$ left unreacted $=0.0416-0.0350=0.0066\text{mol}$
Thus, the mass of $Mg$ left unreacted $=mole\times M_{Mg}=0.0066\times 24=0.16\text{g}$

Hence, the correct answer is option (a).