A 9.00 g pyrolusite sample is fused with KOH and oxidized by air. The overall conversion MnO 2 → K 2 MnO 4 happens with 80% yield (based on MnO 2 actually present). The manganate is fully disp

Question

A 9.00 g pyrolusite sample is fused with KOH and oxidized by air. The overall conversion $MnO_{2} \to$ $K_{2} MnO_{4}$ happens with $80%$ yield (based on $MnO_{2}$ actually present). The manganate is fully disproportionated to $KMnO_{4}$ . The filtered portion of $KMnO_{4}$ dissolved in distilled water to prepare 500 ml aqueous solution The final solution is titrated with 0.200 $M FeSO_{4}$ ; 25.0 mL of ore-derived solution consumes $20.0 mL FeSO_{4}$ in acid. (Given mol. wt. $(MnO_{2}) = 87$ )

Find % purity of $MnO_{2}$

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TestHub · Accepted Answer

Aliquot n ( Fe 2 + ) = 0.200 × 0.0200 = 0.00400 → n ( MnO 4 ​ − ) = 0.00400/5 = 0.000800 Total n ( KMnO 4 ​ ) = 0.000800 × ( 500/25 ) = 0.0160 mol . Disproportionation : n ( MnO 2 ​ ) that actually reacted = 2 3 ​ × 0.01060 = 0.0240 mol Because fusion / air step was 80% yield, moles MnO 2 ​ present = 0.0240/0.80 = 0.030 mol . Mass MnO 2 ​ = 0.030 × 87 = 2.61 Purity = 2.61/9 × 100 = 29.00

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