Question

Assume carbon burns according to following equation :
$2 C _{( s )}+ O _{2( g )} \rightarrow 2 CO ( g )$
When $12 g$ carbon is burnt in $48 g$ of oxygen, the volume of carbon monoxide produced is___ $\times 10^{-1} L$ at STP. [nearest integer]
[Given : Assume $CO$ as ideal gas, Mass of $C$ is $12\, g \,mol ^{-1}$, Mass of $O$ is $16 \,g \,mol ^{-1}$ and molar volume of an ideal gas at STP is $22.7\, L\, mol ^{-1}$ ]

Answer 1

Given a carbon mass of 12 g and a carbon mole of $\frac {12\ g}{12\ g}$= 1 mole.
$\text{Moles of oxygen} = \frac{48g}{32g}\text{ mol}^{−1} = 1.5\text{ mol}$

$2C(s)+O_2(g)\rightarrow2CO(g)$
2mol 1 mol 2 mol
1 mol $\frac{1}{2}$mol 1 mol
In other words, 1 mol carbon will mix with 0.5, 1 mole of oxygen is required to make 1 mole of carbon monoxide.

Therefore, $\text{Mass of 1 mol CO}=28g = \text{Mass of CO produced}$.
Limiting reagent is carbon. One mole carbon produces one mole CO. Hence, volume at STP is $227 \times 10^{-1}$ liter
So, the answer is 227.