Question

Combustion of some amount of ethylene, $6235\text{kJ}$ heat was evolved. The heat of combustion of $1$ mole of ethylene is $1511\text{kJ}$, What will be the volume of $O_2$ (at NTP) that entered into the reaction is?

• A. $277.2\text{mL}$
• B. $277.2\text{litres}$
• C. $6226\text{litres}$
• D. $11.2\text{litres}$

Answer 1

The correct option is A $277.2\text{mL}$

Reaction of Combustion of ethylene is:
$C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(l\right);\Delta H_{\text{comb.}}=1411\text{kJ}$
This $\Delta H_{\text{comb.}}$ is when we have $n_{C_2{H}_4}=1\text{mol}$ or $n_{O_2}=3\text{mol}$, where $n$ represents the number of moles.
So, when $\Delta H_{\text{comb.}}=6235\text{kJ}$
Then moles of oxygen, $n_{O_2}=\left(\frac{6235}{1511}\right)\times 3\text{mol}$

Since the process is taking place at NTP, so the volume of oxygen consumed $=n_{O_2}\times \text{molar volume}$
$=\left(\frac{6235\times 3}{1511}\right)\times 22.4\because \text{molar volume}=22.4\text{L}=277.2\text{litres}$