Q. An impure sample of pyrolusite ore $\left(\right.MnO2\left.\right)$ consist of $70\%MnO_{2}, \, 20\%$ inert impurities and rest is the moisture. On strong heating, all $MnO_{2}$ is converted into MnO alongwith formation of $O_{2}$ . What is the % of Mn in dried sample? (Atomic mass of Mn = 55)

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Solution:

$\left(MnO\right)_{2} & \rightarrow & MnO & + & \frac{1}{2}O_{2} \\ \left(\frac{70}{87}\right) & & \left(\frac{70}{87}\right) & & $
% mass of Mn $=\frac{\left(\frac{70}{87}\right) 55 \times 100}{\left(\frac{70}{87}\right) 71 + 20}=\frac{44 .25}{57 + 20}\times 100=57.4\%$