Question

The reaction,
$2C\left(s\right)+O_2\left(g\right)\to 2CO\left(g\right)$ is carried out by taking 24 g of carbon and 96 g $O_2$. Find out:
(a) Which reactant is left in excess?
(b) How much of it is left?
(c) How many moles of CO are formed?
(d) How many grains of other reactant should be taken so that nothing is left at the end of the reaction?

Answer 1

$\underset{24g}{\underset{2mol}{2C\left(s\right)}}+\underset{32g}{\underset{1mol}{O_2}}\to \underset{56g}{\underset{2mol}{2CO\left(g\right)}}$
Let carbon be completely consumed.
$24g$ carbon give 56 g $CO.$
Let $O_2$ is completely consumed.
$\because$ 32 g $O_2$ give 56 g $CO.$
$\therefore$ 96 g $O_2$ Will give $\frac{56}{32}\times 96gCO=168gCO$
Since, carbon gives least amount of product, te.,56 g $CO$ or 2 mole $CO,$ hence carbon will be the limiting reactant.
$\therefore$ Excess reactant is $O_2$.
Amount of $O_2$ used $=56-24=32g$
Amount of $O_2$ left $=96-32=64g$
$32g$ $O_2$ react with 24 g carbon
$\therefore$ 96 g $O_2$ will react with $72g$ carbon.
Thus, carbon should be taken $72g$ so that nothing is left at the end of the reaction.