Question

Calculate the volume of O2 required
for complete combustion of 12.5 dm3
ethane gas at NTP.

Answer 1

Answer: The volume of oxygen gas required for complete combustion is 31.25 L

Explanation:

At NTP:

1 mole of gas occupies 22.4 L of volume

We are given:

Volume of ethane gas = $12.5dm^3=12.5L$     (Conversion factor:  $1dm^3=1L$ )

The chemical equation for the combustion of ethane gas follows:

$2C_2H_6+5O_2\rightarrow 4CO_2+6H_2O$

By Stoichiometry of the reaction:

$(2\times 22.4)L$ of ethane gas reacts with $(5\times 22.4)L$ of oxygen gas

So, 12.5 L of ethane gas will react with = $\frac{(5\times 22.4)}{(2\times 22.4)}\times 12.5=31.25L$ of oxygen gas

Hence, the volume of oxygen gas required for complete combustion is 31.25 L