Phosphorus burns in oxygen according to the equation P₄ + 50₂ → P₄O₁₀. How many litres of oxygen will be required at S.T.P for complete oxidation of 12.4g of phosphorus? [P = 31,O = 16 and molar volume of a gas at S.T.P = 22.4 litres]

5. 20

11. 20

2. 24

20. 20

6. 20

Correct answer is B

P₄ + 50₂ → P₄O₁₀

for the complete combustion of 124 grammes of phosphorus we need

5 x 22.4 litres of O₂

for 12.4 gms of P₄ we need

$\frac{5 \times 22.4}{124} \times \frac{12.4}{1}$

$\frac{22.4}{2}$ = 11.2 litres

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Phosphorus burns in oxygen according to the equation P₄ + 50₂ → P₄O₁₀. How many litres of oxygen will be required at S.T.P for complete oxidation of 12.4g of phosphorus? [P = 31,O = 16 and molar volume of a gas at S.T.P = 22.4 litres]

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Phosphorus burns in oxygen according to the equation P₄ + 50₂ → P₄O₁₀. How many litres of oxygen will be required at S.T.P for complete oxidation of 12.4g of phosphorus? [P = 31,O = 16 and molar volume of a gas at S.T.P = 22.4 litres]