How much amount of $MgC{O}_3$  in gram having percentage purity 40 percent produces 11.2 L of  $C{O}_2\left(g\right)$ at STP on heating

 $MgC{O}_3\left(s\right)\stackrel{\Delta }{\to }MgO\left(s\right)+C{O}_2\left(g\right)$
    1 mole                     1 mole  $=22.4LatS.T.P$
  $\frac{1}{2}$   mole                      11.2 L
     $=\frac{1}{2}\times 84=42g$
 $m\times 0.40=42$
 $mass=\frac{42}{0.40}=\frac{4200}{40}=105g$