Question

One litre of $c{o}_2$ is passed through red hot coke. The volume becomes $1.4$ litres at same temperature and pressure. The composition of products is:

Answer 1

Chemical Equation:

$C{O}_2+C\to 2CO$

According to the problem, one litre of $C{O}_2$ reacts with 1 mole of Carbon to produce 2 litres of $CO$.

Given 1.4L is the mixtures of gases[CO+unreacted $C{O}_2$].

If $x$ litres of $C{O}_2$ are reacts to give $2x$ litres of $CO$ .

Total volume of mixture will be$=C{O}_2+CO$

Volumes $V$ (mixture) $=V$ (new volume of $C{O}_2$) $+V\left(CO\right)=1.4$ L

$V$(new $C{O}_2$)=V(initial Volume $C{O}_2$)-V(reacted $C{O}_2$) $=(1-x)$ L

$V\left(CO\right)=2x$

So , we get

V(new volume of $C{O}_2$) $+V\left(CO\right)=1.4L$

$(1-x)+2x=1.41-x=1.41-1.4=x$

$X=0.4L$

From, $V$(new $C{O}_2$) $=V$ (initial Volume $C{O}_2$)-V(reacted $C{O}_2$)$=(1-x)$L

$=1-0.4=0.6$ L of $C{O}_2$

Since $V\left(CO\right)=2x$

Volume of $CO=2\times 0.4=0.8L$ of $CO$

Therefore, the composition of product is 0.6 litres of $C{O}_2$ and 0.8 litres of $CO$.