Question

Heat of combustion of $C_2{H}_4$ is $-337K.cal$. If 11.35 lit $O_2$ is used at STP, in the combustion heat liberated is ___________ $K.cal$.

• A. 56.17
• B. 14.04
• C. 42.06
• D. 27.7

Answer 1

The correct option is A 56.17

Solution:- (A) $56.17\text{kcal}$

Combustion of $C_2{H}_4\to$

${C_2{H}_4}_{\left(g\right)}+3{O_2}_{\left(g\right)}⟶2{C{O}_2}_{\left(g\right)}+2{H_{2}O}_{\left(l\right)};\Delta H=-337kcal$

At STP,

$1$ mole of a gas $=22.4L$

$\Rightarrow 3$ moles of a gas $=3\times 22.4=67.2L$

Given amount of oxygen $=11.35L$

Now from the above reaction,

Heat liberated when $67.2L$ of oxygen is used $=337kcal$

Heat liberated when $11.35L$ of oxygen is used $=\frac{337}{67.2}\times 11.35=56.91\approx 56.17\text{kcal}$

Hence the amount of heat liberated when $11.35L$ of oxygen is used is $56.17\text{kcal}$.