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

The correct option is B $84$
$MgC{O}_3\left(s\right)\to MgO\left(s\right)+C{O}_2\left(g\right)$
Moles of $MgC{O}_3=\frac{20}{84}\text{mol}$
From the above equation
1 mole of $MgC{O}_3$ gives 1 mole $MgO$
$\frac{20}{84}\times 40\text{g}=\frac{200}{21}\text{g}MgO$
Practical yield of $MgO=8\text{g}MgO$
$\therefore$ % purity =$\frac{8\times 21}{200}\times 100=84\%$