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1.6 g of pyro...

Question

$1.6 g$ of pyrolusite was treated with $60 m L$ of normal oxalic acid and some $H_{2} S O_{4}$ . The oxalic acid left undecomposed was made up to $250 m L, 25 m L$ of this solution required $32 m L$ of $0.1 N$ potassium permanganate $(K M n O_{4})$ . Calculate the percentage of pure $M n O_{2}$ in pyrolusite.

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Solution

m.eq of "excess of oxalic acid" = $32 \times 0.1$ in $25 m l$
For $250 m L$ = $32 \times 0.1 \times \frac{250}{25} = 32 m e q$

Total oxalic acid used = $50 \times 1 = 50 m e q$

m. Eq of "oxalic acid used for $M n O_{2} = 50 - 32 = 18$
m. Eq of $M n O_{2} = 18$
Mass of $M n O_{2} = 18 \times 10^{- 3} \times \frac{87}{2} g o f M n O_{2}$ = $0.783 g$

% of $M n O_{2} = \frac{0.783}{1.6} \times 100 = 48.93$ % [Answer]

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