Question

Magnesium sulphide was obtained from $4.00\text{g}$ of magnesium and $2.00\text{g}$ of sulphur by the reaction $Mg+S\to MgS$. The volume of oxygen at STP ($1atm,273K$) required for the complete oxidation of the excess reagent is:

• A. 0.93 L
• B. 0.53 L
• C. 0.63 L
• D. 1.3 L

Answer 1

The correct option is A 0.93 L

moles of $Mg=$ $\frac{2}{24}=\frac{1}{12}$
moles of $S=$ $\frac{4}{32}=\frac{1}{8}$

Finding the limiting reagent,
$\frac{\text{moles}}{\text{stoichiometric coefficient}}$

$\frac{\text{moles of Mg}}{1}=\frac{1}{12}$
$\frac{\text{moles of S}}{1}=\frac{1}{8}$
Hence $Mg$ is the limiting reagent.
Excess moles of $S$ left = $\frac{1}{8}-\frac{1}{12}=\frac{1}{24}$
oxidation of S,
$S+O_2\to S{O}_2$

By unitary method,
$\frac{\text{moles of S}}{1}=\frac{\text{moles of}{O}_2}{1}$
Moles of oxygen = $\frac{1\times 1}{24\times 1}=\frac{1}{24}$
Volume of oxygen at STP = $22.4\times \frac{1}{24}=0.93L$