Question

How many litres of oxygen (at STP) are required for complete combustion of 39 grams of liquid benzene​

Answer 1

84

The combustion reaction is : C 6 H 6 + 15 2 O 2 → 6 C O 2 + 3 H 2 O This means that 15 × 22.4 2 L of O 2 at STP is required for 78 grams of benzene. (Mol. wt of Benzene = 78 ) For 39 grams of benzene, 39 × 15 × 22.4 78 × 2 = 84 grams of O 2 is required.