The volume of $C{O}_2$ that can be obtained at STP from 60g. of 70% pure $MgC{O}_3$ is

$\text{ Weight of pure }{\mathrm{MgCO}}_3=\frac{70}{100}\times 60=42g$   ${\mathrm{MgCO}}_3\to \mathrm{MgO}={\mathrm{CO}}_2$

$84g------\underset{\underset{22.4 L}{\downarrow }}{44g}$                                             $42g--------11.2L$