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Standard XII

Chemistry

Order of Reaction

0.261 g of a ...

Question

$0.261 g$ of a sample of pyrolusite was heated with an excess of $H C l$ and the chlorine evolved was passed in a solution of $K I$ . The liberated iodine required $90 m L \frac{N}{30} N a_{2} S_{2} O_{3}$ . Calculate the percentage of $M n O_{2}$ in the sample.

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Solution

$M n O_{3} + 4 H C l \to M n C l_{2} + 2 H_{2} O + C l_{2}$ $
$2 K I + C l_{2} \to 2 K C l + I_{2}$
$2 N a_{2} S_{2} O_{3} + I_{2} \to N a_{2} S_{4} O_{6} + 2 N a I$
$90 m L \frac{N}{30} N a_{2} S_{2} O_{3} \equiv 90 m L \frac{N}{30} I_{2}$
$\equiv 90 m L \frac{N}{30} C l_{2}$
$\equiv 90 m L \frac{N}{30} M n O_{2}$
Eq. mass of $M n O_{2} = \frac{M o l . m a s s}{2} = \frac{87}{2}$
[Since, change in O.N. is from $4$ to $2$ ]
Amount of $M n O_{2} = \frac{87}{2 \times 30} \times \frac{90}{1000} = 0.1305 g$
% of $M n O_{2} = \frac{0.1305}{0.261} \times 100 = 50$ .

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0.261 g of a sample of pyrolusite was heated with an excess of HCl and the chlorine evolved was passed in a solution of KI. The liberated iodine required 90 mL N30Na2S2O3. Calculate the percentage of MnO2 in the sample.

MnO3+4HCl→MnCl2+2H2O+Cl2$2KI+Cl2→2KCl+I22Na2S2O3+I2→Na2S4O6+2NaI90 mLN30Na2S2O3≡90 mLN30I2≡90 mLN30Cl2≡90 mLN30MnO2Eq. mass of MnO2=Mol.mass2=872[Since, change in O.N. is from 4 to 2]Amount of MnO2=872×30×901000=0.1305 g% of MnO2=0.13050.261×100=50.

$0.261 g$ of a sample of pyrolusite was heated with an excess of $H C l$ and the chlorine evolved was passed in a solution of $K I$ . The liberated iodine required $90 m L \frac{N}{30} N a_{2} S_{2} O_{3}$ . Calculate the percentage of $M n O_{2}$ in the sample.