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How many grams of pure $K M n O_{4}$ should be heated with $H_{2} S O_{4}$ to produce 48 grams $O_{2}$ ?
$K M n O_{4} + H_{2} S O_{4} \to K_{2} S O_{4} + M n S O_{4} + H_{2} O + O_{2}$
$[M n = 55, K = 39, O = 16, H = 1, S = 32]$

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answer is 189.6.

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Detailed Solution

The balanced reaction is $4 K M n O_{4} + 6 H_{2} S O_{4} \to 2 K_{2} S O_{4} + 4 M n S O_{4} + 6 H_{2} O + 5 O_{2}$
$\frac{n K M n O_{4}}{n O_{2}} = \frac{4}{5}$
Wt. of $\frac{n K M n O_{4}}{n O_{2}} = \frac{4}{5}$
=189.6 gram

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