Question

When a certain amount of ethylene was combusted, 6226 kJ heat was evolved. If heat of combustion of ethylene is 1411 kJ, the volume of O2 (at NTP) that entered into the reaction is:

• A. 296.5 ml
• B. 296.3 L
• C. 6226 × 22.4 L
• D. 22.4 L

Answer 1

Answer: (B)

296.3 L

Answer 2

$n_{C_{2}H_{4}} = \frac{6226}{1411}$= 4.41

C2H4 + 3O2 $\longrightarrow$ 2CO2 + 2H2O

4.41

$V_{O_{2}}$ = 4.41 × 3 × 22.4 = 296.3 L