Question

What volume of oxygen gas $\left(O_2\right)$ measured at $0^{\circ }\text{C}$ and $1\text{atm}$ is needed to burn $500\text{mL}$ of propane gas $\left(C_3{H}_8\right)$ completely, measured under the same conditions?

• A. $10\text{L}$
• B. $7.5\text{L}$
• C. $2.5\text{L}$
• D. $5\text{L}$

Answer 1

The correct option is C $2.5\text{L}$

$C_3{H}_8+5O_2\to 3C{O}_2+4H_{2}O$
$1\text{mol}$ of $C_3{H}_8$ reacts with $5\text{mol}$ of $O_2$
i.e., $22.4\text{L}$ of $C_3{H}_8$ reacts with $5\times 22.4\text{L of}{O}_2$
So, $500\text{mL}$ of $C_3{H}_8$ will react with $\left(\frac{5\times 22.4\times 0.5}{22.4}\right)=2.5\text{L of}{O}_2$