Question

$1 \,g$ of $Mg$ is burnt in a closed vessel containing $0.5\, g$ 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 $0.25 \,g$
• B. $Mg$ is a limiting reagent and is in excess by $0.5\, g$
• C. $O_{2}$ is a limiting reagent and is in excess by $0.25 \,g$
• D. $O_{2}$ is a limiting reagent and $Mg$ is in excess by $0.75 \,g$

Answer 1

Answer: (A)

$\underset{2 \times 24}{2 Mg} + \underset{2 \times 16}{O _2} \longrightarrow \underset{2(24+16)}{2 MgO}$
$48\,g$ of $Mg$ requires $32\,g$ of $O_{2}$
$1\, g$ of $Mg$ requires $\frac{32}{48} = 0.66\,g$ of $O_{2}$
Oxygen available $= 0.5\, g$
Hence, $O_{2}$ is limiting reagent,
$ 32 \,g$ of $O_ {2}$ reacts with $48 \,g$ of $Mg$
$0.5\, g$ of $O_{2}$ will react with $\frac{48}{32} \times 0.5 = 0.75\,g$ of $Mg$
Excess of $Mg = (1.0-0.75) = 0.25\,g$