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Q. An impure sample of pyrolusite ore $\left(\right.MnO_2\left.\right)$ consist of $70\%MnO_{2}, \, 20\%$ inert impurities and rest is the moisture. On strong heating, all $MnO_{2}$ is converted into $MnO$ along with formation of $O_{2}$ . What is the $\%$ of $Mn$ in dried sample? (Atomic mass of $Mn = 55)$

NTA AbhyasNTA Abhyas 2022

Solution:

$\underset{\frac{70}{87}}{MnO_2} \rightarrow \underset{\frac{70}{87}}{MnO} + \frac{1}{2} O_2$
% mass of $Mn =\frac{\left(\frac{70}{87}\right) 55 \times 100}{\left(\frac{70}{87}\right) 71 + 20}=\frac{44 .25}{57 + 20}\times 100=57.4\%$