Question

$896mL$ of a mixture of $CO$ and $C{O}_2$ weight $1.28g$ at $NTP$. Calculate the volume of $C{O}_2$ in the mixture at $NTP$.

Answer 1

Let the weight of $CO$ in the mixture be x g.

Total weight of mixture $=1.28g$

Weight of $C{O}_2=1.28-x$

Moleculr weight of $CO=28g$

Molecular weight of $C{O}_2=44g$

Moles of $C{O}_2$ in the mixture $=\frac{1.28-x}{44}$

Moles of $CO$ in the mixture $=\frac{x}{28}$

At NTP, the volume occupied by 1 mole of a gas $=22400mL$

$\therefore$ 896 mL occupied by $\frac{1}{22400}\times 896=0.04\text{moles}$

$\therefore$ Total moles in the mixture = 0.04 moles

$\Rightarrow$ Moles of $C{O}_2+$ Moles of $CO=0.04$

$\Rightarrow \frac{1.28-x}{44}+\frac{x}{28}=0.04$

$\Rightarrow 35.84-28x+44x=49.28$

$\Rightarrow 16x=13.44$

$\Rightarrow x=0.84$

$\therefore$ Molecular weight of $C{O}_2=1.28-0.84=0.44g$

$\therefore$ Moles of $C{O}_2$ in the mixture $=\frac{0.44}{44}=0.01\text{moles}$

$\because$ Volume occupied by 1 mole of $C{O}_2=22400mL$

$\Rightarrow$ Volume occupied by 0.01 mole of $C{O}_2=22400\times 0.01=224mL$

Hence, at NTP, 224 mL of $C{O}_2$ is present in the mixture.