How many grams of ${C l}_{2}$ gas will be obtained by the complete reaction of $31.5 g m$ of potassium permanganate with hydrochloric acid?

[Molar mass of $K M n O_{4} = 316 g m / m o l$ ]

71

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17.75

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35.5

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142

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Solution

The correct option is B $17.75$
$2 K M n O_{4} + 16 H C l \to 2 K C l + 2 M n C l_{2} + 8 H_{2} O + 5 C l_{2}$

Moles of $31.5 g m$ of potassium permanganate $= \frac{31.5 g m}{316 g m / m o l} = 0.0996$ mole.

0.0996 moles of potassium permanganate will give $\frac{5}{2} \times 0.0996 = 0.249$ moles of chlorine.

The molar mass of $C l_{2} = 71$ g/mole.

The mass of $C l_{2} = 71$ g/mole $\times 0.249$ mol $= 17.75$ g.

Hence, option B is correct.

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How many grams of Cl2 gas will be obtained by the complete reaction of 31.5 gm of potassium permanganate with hydrochloric acid?[Molar mass of KMnO4=316 gm/mol]

How many grams of ${C l}_{2}$ gas will be obtained by the complete reaction of $31.5 g m$ of potassium permanganate with hydrochloric acid?

[Molar mass of $K M n O_{4} = 316 g m / m o l$ ]