Question

The volume of oxygen at $STP$ required to burn $2.4g$ of carbon completely is

• A. $1.12L$
• B. $8.96L$
• C. $2.24L$
• D. $4.48L$

Answer 1

Answer: (D)

$C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right)$
1 mole of carbon reacts completely with
1 mole of oxygen to produce 1 mole of $C{O}_2$.
$12g$ of $C$ reacts = lmole of $O_2$
$2.4g$ of $C$ reacts with $=\frac{1}{12}\times 2.4$
$=0.2$ mole of $O_2$
At $STP1$ mole of $O_2$ contains $=22.4L$
$\therefore 0.2$ mole of $O_2$ contains $=22.4\times 0.2=4.48L$