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Standard XII

Chemistry

Introduction

3.0 g of pyro...

Question

$3.0 g$ of pyrolusite ore were treated with $20 g$ of pure ferrous ammonium sulphate $(M o l . m a s s = 392 g m o l^{- 1})$ and dilute $H_{2} S O_{4}$ . After the reaction, the solution was diluted to $500 m L$ . $50 m L$ of diluted solution required $10 m L$ of $0.1 N K_{2} C r_{2} O_{7}$ solution. Calculate the % of pure $M n O_{2}$ in pyrolusite.

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Solution

$M n O_{2}$ present in pyrolusite oxidises ferrous ammonium sulphate into ferric ammonium sulphate, i.e., $F e^{2 +} \to F e^{3 +}$ .
Unreacted ferrous ammonium sulphate is estimated by potassium dischromate solution.
$M n O_{2} + 2 F e^{2 +} + 4 H^{+} \to M^{2 +} + 2 H_{2} O + 2 F e^{3 +}$
$C r_{2} O_{7}^{2 +} + 6 F e^{2 +} + 14 H^{+} \to 2 C r_{3 +} + 7 H_{2} O + 6 F e^{3 +}$
$50 m L$ diluted ferrous ammonium sulphate solution
$= 10 m L$ of $0.1 N K_{2} C r_{2} O_{7}$
$500 m L$ diluted ferrous ammonium sulphate solution
$= 10 \times 10 m L$ of $0.1 N K_{2} C r_{2} O_{7}$
$= 100 m L$ of $0.1 N F e A S$
$= \frac{0.1 \times 392}{1000} \times 100 = 3.92 g$
Used $F e A S = (20 - 3.92) = 16.08 g$

$M n O_{2}$ persent in pyrolusite $= \frac{87}{392 \times 2} \times 16.08 = 1.784 g$

Percentage of pure $M n O_{2} = \frac{1.784}{3.0} \times 100 = 59.4$ %.

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Similar questions

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

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32 m L

0.1 N

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Q. A

3.0 g

sample containing

F e_{3} O_{4}, F e_{2} O_{3}

and an inert impure substance is treated with excess of

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0.5 M N a_{2} S_{2} O_{3}

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50 m L

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Q. 4.35 g of pyrolusite (

M n O_{2}

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1.6 g

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3.0 g of pyrolusite ore were treated with 20 g of pure ferrous ammonium sulphate (Mol.mass=392 g mol−1) and dilute H2SO4. After the reaction, the solution was diluted to 500 mL. 50 mL of diluted solution required 10 mL of 0.1 N K2Cr2O7 solution. Calculate the % of pure MnO2 in pyrolusite.