Question

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

• A. $5L$
• B. $10L$
• C. $7L$
• D. $6L$

Answer 1

Answer: (A)

$5L$

Volume of propane $=5L$

$\therefore 5O_2+C_3{H}_8\to 3C{O}_2+4H_{2}O$

From the balanced reaction we can get,

Volume of $O_2=$ Volume of $C_3{H}_8\times 5$

Since $V\propto n$ ------ [At constant T and P]

Volume of $O_2=1\times 5=5L$