16.8 grams of Magnesite on calcination liberates 3.36 litres of ${\mathrm{CO}}_2$ at STP.  Percentage Purity of Magnesite is

$\begin{array}{l}84\mathrm{g}22.4\mathrm{lit}\left(\mathrm{STP}\right)\\ {\mathrm{MgCO}}_{3\left(\mathrm{s}\right)}\to {\mathrm{MgO}}_{\left(\mathrm{s}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}\\ 16.8\mathrm{g}\text{'}\mathrm{X}\text{'} \text{\hspace{0.17em}lit}\left(\mathrm{STP}\right)\end{array}$
$\mathrm{X}=\left(\frac{16.8}{84}\times 22.4\right)\mathrm{lit}$
X=4.48l
But only 3.36l of  ${\mathrm{CO}}_2$ at STP is liberated
$\therefore \mathrm{Purity} \mathrm{of} \mathrm{Magnesite} =\frac{3.36}{4.48}\times 100=75\mathrm{\%}$