Question

What would be percentage composition by volume of a mixture of CO and CH4, whose 10.5 ml requires 9 ml oxygen for complete combustion?

Answer 1

Answer: The volume percent of CO and methane gas in the mixture is 76.2 % and 23.8 % respectively.

Explanation:

We are given:

Volume of mixture = 10.5 mL

The chemical equation for the combustion of CO follows:

$2CO+O_2\rightarrow 2CO_2$

The chemical equation for the combustion of methane follows:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

Let the volume of CO in the mixture be 'x' mL and the volume of methane in the mixture be (10.5-x) mL

• For CO:

By Stoichiometry of the reaction:

2 mL of CO requires 1 mL of oxygen gas

So, 'x' mL of CO will require = $\frac{1}{2}\times x=\frac{x}{2}mL$ of oxygen gas

Volume of oxygen gas required for combustion of CO = $\frac{x}{2}mL$

• For methane:

By Stoichiometry of the reaction:

1 mL of methane requires 2 mL of oxygen gas

So, (10.5-x) mL of methane will require = $\frac{2}{1}\times (10.5-x)=(21-2x)mL$ of oxygen gas

Volume of oxygen gas required for combustion of methane = (21-2x) mL

We are given:

Total volume of oxygen gas = 9 mL

$\frac{x}{2}+(21-2x)=9\\\\42-3x=18\\\\x=8$

To calculate the volume percentage, we use the equation:

$\text{Volume percent of substance}=\frac{\text{Volume of substance}}{\text{Volume of mixture}}\times 100$       .......(1)

Volume of mixture = 10.5 mL

• For CO:

Volume of CO = 8 mL

Putting values in equation 1, we get:

$\text{Volume percent of CO}=\frac{8mL}{10.5mL}\times 100=76.2\%$

• For methane:

Volume of methane = (10.5 - 8) = 2.5 mL

Putting values in equation 1, we get:

$\text{Volume percent of }CH_4=\frac{2.5mL}{10.5mL}\times 100=23.8\%$

Hence, the volume percent of CO and methane gas in the mixture is 76.2 % and 23.8 % respectively.

Answer 2

Answer:

23.8% and 75.2%

Explanation:

View the image for the explanation