Question

$20.0g$ of a magnesium carbonate sample decomposes on heating to give carbon dioxide and $8.0g$ of magnesium oxide. What will be the percentage purity of magnesium carbonate in the sample? (At. Wt. $Mg=24$)

• A. 60
• B. 84
• C. 75
• D. 96

Answer 1

Answer: (B)

Correct option: (B)

$\text{Percentage purity}=\frac{\text{Mass of magnesium carbonate actually present}}{\text{Mass of magnesium carbonate that should be theoretically present}}$
The decomposition reaction of $MgC{O}_3$ can be written as:

$MgC{O}_3\to MgO+C{O}_2$

From stoichiometry, 1 mole of $MgC{O}_3$ decomposes to produce 1 mole of $MgO$

Moles of $MgO$ formed $=\frac{Givenmass}{Molarmass}=\frac{8}{40}=0.2mol$

Hence, 0.2 mol of $MgC{O}_3$ reacted.

% purity $=\frac{84}{20}\times 0.2\times 100=84$ %