When heated, potassium permanganate decomposes according to the following equation:
$2 K M n O_{4} \to K_{2} M n O_{4} + M n O_{2} + O_{2}$
Given that the molecular mass of potassium permanganate is $158$ , what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of $15.8$ g of potassium permanganate? (Molar volume at room temperature is $24$ liters.)

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Solution

Molar mass of $K M n O_{4}$ $= 158$
Molar volume of $O_{2}$ at room temperature $= 24$ litre
$158$ g of $K M n O_{4}$ at room temperature yields $24$ litres of $O_{2}$
$15.8$ g of $K M n O_{4}$ at room temp. will yield $= \frac{24}{2 \times 158} \times 15.8 = 1.2$ litres of $O_{2}$ .

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When heated, potassium permanganate decomposes according to the following equation:

Given that the molecular mass of potassium permanganate is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres.)

When heated, potassium permanganate decomposes according to the following equation:
2KMnO₄ $\overset{}{\to}$ K₂MnO₄ + MnO₂ + O₂
(a) Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.
(b) Given that the molecular mass of potassium permanganate is 158 g, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate? (Molar volume at room temperature is 24 litres). (K = 39, Mn = 55, O = 16)

When heated, potassium permanganate decomposes according to the following equation:

Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same conditions of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen.

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Why potassium permanganate discolouruse faster at higher temperature than that at lower temperature?

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Molar mass of KMnO4 =158Molar volume of O2 at room temperature =24 litre158 g of KMnO4 at room temperature yields 24 litres of O215.8 g of KMnO4 at room temp. will yield =242×158×15.8=1.2 litres of O2.

Molar mass of $K M n O_{4}$ $= 158$
Molar volume of $O_{2}$ at room temperature $= 24$ litre
$158$ g of $K M n O_{4}$ at room temperature yields $24$ litres of $O_{2}$
$15.8$ g of $K M n O_{4}$ at room temp. will yield $= \frac{24}{2 \times 158} \times 15.8 = 1.2$ litres of $O_{2}$ .