A sample of $MgC{O}_3$  is 50% pure if on heating $4.0gm$ of $MgO$ and $4.4gm$  of $C{O}_2$ are formed. Calculate the amount of $MgC{O}_3$ taken.

$$\begin{gathered} \underset{84gm}{MgC{O}_3}\stackrel{\Delta }{\to }\underset{40gm}{MgO}+\underset{44gm}{C{O}_2} \\ x\leftarrow 4.0gm4.4gm \\ x=8.4gm \\ But MgC{O}_3 is 50\% pure. Hence amount of MgC{O}_3 taken =8.4\times \frac{100}{50}=16.8gm \end{gathered}$$