Question

1 g 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 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 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)

$O_{2}$is a limiting reagent and Mg is in excess by $0.25g$

Answer 2

(1) : $\underset{2 \times 24}{2Mg} + \underset{2 \times 16}{O_{2}} \rightarrow \underset{2(24 + 16)}{2MgO}$

48 g of Mg requires 32 g of $O_{2}$

1 g of Mg requires $\frac{32}{48} = 0.66g$of $O_{2}$

Oxygen available = 0.5 g

Hence, $O_{2}$is limiting reagent.

32 g of $O_{2}$reacts with 48 g Mg

$O_{2}$will react with $\frac{48}{32} \times 0.5 = 075g$of Mg

Excess of Mg

= $(1.0 - 0.75) = 0.25g$