Question

If $12.0$ L of $H_2$ and $8.0$ L of $O_2$ are allowed to react, the unreacted $O_2$ left will be:

• A. $4.0$ L
• B. $6.0$ L
• C. $1.0$ L
• D. $2.0$ L

Answer 1

The correct option is D $2.0$ L

The reaction is $H_2+\frac{1}{2}{O}_2\to H_{2}O$.

At STP conditions, one mole of any gas will occupy $22.4$ L.

$12.0$ L of hydrogen and $8.0$ L of oxygen will correspond to $\frac{12.0}{22.4}=0.54$ moles and $\frac{8.0}{22.4}=0.36$ moles respectively.

One mole of hydrogen reacts with half mole of oxygen.

Hence, $0.54$ moles of hydrogen will react with $0.27$ moles of oxygen.

Hence, the number of moles of oxygen unreacted will be $0.36-0.27=0.09$ moles.

The volume of oxygen unreacted will be $0.09\times 22.4=2.0$ L.

Hence, option D is correct.