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Mole Concept

MnO2 on ignit...

Question

$M n O_{2}$ on ignition converts into $M n_{3} O_{4}$ . A sample of pyrolusite having $75 % M n O_{2}, 20 %$ inert impurities and rest water is ignited in air to constant weight. What is the percentage of $M n$ in the ignited sample?

24.6 %

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37 %

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55.24 %

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74.05 %

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Solution

The correct option is C $55.24 %$
Let the weight of sample be $x g$ . Then, weight of $M n O_{2}$ is $0.75 x g$ .
Moles of $M n O_{2} = \frac{0.75 x}{87}$ .
Weight of $M n = \frac{0.75 x}{87} \times 55 = 0.474 x$ .
$3 M n O_{2} \to M n_{3} O_{4} + O_{2}$ .
Weight of $M n_{3} O_{4}$ produced $= \frac{0.75 x}{87} \times \frac{1}{3} \times 229 = 0.658 x g$ .
Total weight of residue $= 0.658 x + 0.2 x$ (weight of impurities).
$%$ of $M n$ in residue $= \frac{0.474 x}{0.858 x} \times 100 = 55.24$ .

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MnO2 on ignition converts into Mn3O4. A sample of pyrolusite having 75%MnO2,20% inert impurities and rest water is ignited in air to constant weight. What is the percentage of Mn in the ignited sample?

$M n O_{2}$ on ignition converts into $M n_{3} O_{4}$ . A sample of pyrolusite having $75 % M n O_{2}, 20 %$ inert impurities and rest water is ignited in air to constant weight. What is the percentage of $M n$ in the ignited sample?