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Stoichiometric Calculations

0.5 g sample ...

Question

$0.5$ g sample containing $M n O_{2}$ is treated with $H C l$ liberating $C l_{2}$ . The $C l_{2}$ is passed into a solution of $K I$ and $30.0$ cc of $0.1 M N a_{2} S_{2} O_{3}$ are required to titrate the liberated iodine. Calculate the percentage of $M n O_{2}$ in sample (atomic mass of $M n = 55$ ).

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Solution

The complete reaction that occurs is given below:
$M n O_{2} H C l - - \to C l_{2} K I - - \to I_{2} N a_{2} S_{2} O_{3} - ---- \to N a I + N a_{2} S_{4} O_{6}$
Redox changes are as follows:
$2 e + I_{2}^{0} ⟶ 2 I^{-}$
$2 (S^{2 +})_{2} ⟶ {⎛ ⎜ ⎝ S^{\frac{5}{2} +} ⎞ ⎟ ⎠}_{4}^{\frac{5}{2} +} + 2 e$
$2 e + M n^{4 +} ⟶ M n^{2 +}$

$[∵ N_{N a_{2} S_{2} O_{3}} = M_{N a_{2} S_{2} O_{3}}$ because the valence factor $= 1]$

Meq. of $M n O_{2} =$ Meq. of $C l_{2}$ formed $=$ Meq. of $I_{2}$ liberated $=$ Meq. of $N a_{2} S_{2} O_{3}$ used
$∴ \frac{w}{(\frac{M}{2})} \times 1000 = 0.1 \times 1 \times 30$

$w = \frac{0.1 \times 1 \times 30 \times M}{2000} = \frac{0.1 \times 1 \times 30 \times 87}{2000}$ $(∵ M_{M n O_{2}} = 87)$

$w_{M n O_{2}} = 0.1305$
Purity of $M n O_{2} = \frac{0.1305}{0.5} \times = 26.1 %$
So, the nearest integer value is $26$ .

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Q. A

0.5

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0.5 g sample containing MnO2 is treated with HCl liberating Cl2. The Cl2 is passed into a solution of KI and 30.0 cc of 0.1MNa2S2O3 are required to titrate the liberated iodine. Calculate the percentage of MnO2 in sample (atomic mass of Mn=55).