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Oxidation Number Method

A quantity of...

Question

A quantity of $4.35 g$ of a sample of pyrolusite ore, when heated with conc. $H C l$ , gave chlorine. The chlorine when passed through potassium iodide solution, liberated $6.35 g$ of iodine. The percentage of pure $M n O_{2}$ in the pyrolusite ore is: $(M n = 55, I = 127)$ .

40

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Solution

The correct option is B $50$
Given, Pyrolusite means $M n O_{2}$ ore
The reaction involved is
$M n O_{2} + 4 H C l ⟶ M n C l_{2} + C l_{2} + 2 H_{2} O$
Again, given that this chlorine is passed through $K I$
$C l_{2} + 2 K I ⟶ 2 K C l + I_{2}$
Here $I_{2}$ liberated= $6.35 g$
Using the above equations it is possible to find mass of $M n O_{2}$
$\Rightarrow$ No. of moles of $I_{2}$ liberated= $\frac{6.35}{254 g / m o l} = 0.025$ mol
From abore equation
$1$ mole $M n O_{2} ⟶ 1$ mol $C l_{2}$
& also $1$ mole $C l_{2} ⟶ 1$ mole $I_{2}$
that means $0.025$ mol $I_{2}$ came from $0.025$ mol $M n O_{2}$ initially present.
$∴$ In given sample
Mass of $M n O_{2}$ present= $0.025 m o l \times 87$ $g / m o l$
$= 2.175 g$
$∴$ Purity%= $\frac{2.175}{T o t a l m a s s o f s a m p l e} \times 100$
$= \frac{2.175 g}{4.35 g} \times 100 = 50$ %

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A quantity of 4.35 g of a sample of pyrolusite ore, when heated with conc. HCl, gave chlorine. The chlorine when passed through potassium iodide solution, liberated 6.35 g of iodine. The percentage of pure MnO2 in the pyrolusite ore is: (Mn=55,I=127).

A quantity of $4.35 g$ of a sample of pyrolusite ore, when heated with conc. $H C l$ , gave chlorine. The chlorine when passed through potassium iodide solution, liberated $6.35 g$ of iodine. The percentage of pure $M n O_{2}$ in the pyrolusite ore is: $(M n = 55, I = 127)$ .