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]
A. 5. 20 B. 11. 20 C. 2. 24 D. 20. 20 E. 6. 20
Correct Answer: B
Explanation
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
