If heat of combustion of enthylene is 1411 KJ when a certain amount of ethylene was burnt 6226 KJ heat was evolved. Then the volume of `O_(2)` (at NTP) that entered into the reaction is :-

Options

A. 296.5 mL
B. 296.5 litre
C. `6226 xx 22.4 ` litre
D. 22.4 litre

Answer

B

Explanation

`C_(2)H_(4)+3O_(2)rarr2CO_(2)+2H_(2)O`
`underset("(1 mole )")(C_(2)H_(4)) -= underset("( 3 mole )")(3O_(2)) " " DeltaH=-1411 " kJ"`
`(3xx 22.4) L`
`therefore V_(O_2)` used `= (-6226 xx 3 xx 2.4)/(-1411) = 296.5` litre