Question

$1g$ of $Mg$ is burnt in a closed vessel containing $0.5g$ of $O_2$. Which reactant is limiting reagent and how much of the excess reactant will be left?

• A. $O_2$ is a limiting reagent and $Mg$ is in excess by $0.25g$
• B. $Mg$ is a limiting reagent and is in excess by $0.5g$
• C. $O_2$ is a limiting reagent and is in excess by $0.25g$
• D. $O_2$ is a limiting reagent and $Mg$ is in excess by $0.75g$

Answer 1

Answer: (A)

$\underset{2\times 24}{2Mg}+\underset{2\times 16}{O_2}⟶\underset{2(24+16)}{2MgO}$
$48g$ of $Mg$ requires $32g$ of $O_2$
$1g$ of $Mg$ requires $\frac{32}{48}=0.66g$ of $O_2$
Oxygen available $=0.5g$
Hence, $O_2$ is limiting reagent,
$32g$ of $O_2$ reacts with $48g$ of $Mg$
$0.5g$ of $O_2$ will react with $\frac{48}{32}\times 0.5=0.75g$ of $Mg$
Excess of $Mg=(1.0-0.75)=0.25g$