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
The correct answer is: B
Explanation
P4 + 502 → P4O10for 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