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Standard XII

Chemistry

Introduction

4.35 g of pyr...

Question

4.35 g of pyrolusite ( $M n O_{2}$ mineral) is digested with conc. $H C l$ and $K I$ added. Mixture required 5 mL of 1 M $N a_{2} S_{2} O_{3}$ solution for complete reaction of $K I_{3}$ formed. What is the $%$ of pure $M n O_{2}$ in pyrolusite? $(M n = 55)$

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Solution

Let the percentage of pure $M n O_{2}$ in pyrolusite be X.
The chemicalreactions are as given below.
$2 K I + 4 H C l + M n O_{2} \to 2 K C l + M n C l_{2} + 2 H_{2} O + I_{2}$
$K I + I_{2} ⇌ K I_{3}$
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
Thus, it can be seen that 1 mole of $M n O_{2}$ reacts with 2 moles of $N a_{2} S_{2} O_{3}$
the number of moles of $N a_{2} S_{2} O_{3}$ present is $1 m o l / L \times 0.005 L = 0.005 m o l e s$

hence, the number of moles of $M n O_{2}$ present is $\frac{0.005 m o l e s}{2} = 0.0025 m o l e s$

This corresponds to $0.0025 m o l e s \times 87 g / m o l = 0.2175 g$

Hence, the percentage purity of $M n O_{2}$ is $\frac{0.2175}{4.35} \times 100 = 5$ %

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4.35 g of pyrolusite (MnO2 mineral) is digested with conc. HCl and KI added. Mixture required 5 mL of 1 M Na2S2O3 solution for complete reaction of KI3 formed. What is the % of pure MnO2 in pyrolusite? (Mn=55)

4.35 g of pyrolusite ( $M n O_{2}$ mineral) is digested with conc. $H C l$ and $K I$ added. Mixture required 5 mL of 1 M $N a_{2} S_{2} O_{3}$ solution for complete reaction of $K I_{3}$ formed. What is the $%$ of pure $M n O_{2}$ in pyrolusite? $(M n = 55)$