In the equation CS_2(l) + 3O_2(g) -> CO_2(g) + 2SO_2(g), if 1.00 mol of CS_2 is combined with 1.00 mol of O_2, what is the limiting reactant?

Mar 20, 2016

Oxygen gas, ${\text{O}}_{2}$.

Explanation:

Take a look at the balanced chemical equation for this reaction

${\text{CS"_text(2(l]) + color(red)(3)"O"_text(2(g]) -> "CO"_text(2(g]) + 2"SO}}_{\textrm{2 \left(g\right]}}$

As you can see, carbon disulfide, ${\text{CS}}_{2}$, will react with oxygen gas, ${\text{O}}_{2}$, in a $1 : \textcolor{red}{3}$ mole ratio.

This tells you that in order for the reaction to take place, you need to have $\textcolor{red}{3}$ moles of oxygen gas for every $1$ mole of carbon disulfide.

In other words, every mole of carbon disulfide will consume $\textcolor{red}{3}$ moles of oxygen gas.

Notice that your reaction uses $1.00$ mole of carbon disulfide, so right from the start you know that in order for all the carbon disulfide to react, you need to have

1.00color(red)(cancel(color(black)("mole CS"_2))) * (color(red)(3)color(white)(a)"moles O"_2)/(1color(red)(cancel(color(black)("mole CS"_2)))) = "3.00 moles O"_2

However, you only have $1.00$ mole of oxygen gas available. This means that oxygen gas will act as a limiting reagent, i.e. it will determine how much carbon disulfide actually reacts and how much remains in excess.

To see how much carbon disulfide will react, use the same $1 : \textcolor{red}{3}$ mole ratio

1.00color(red)(cancel(color(black)("mole O"_2))) * "1 mole CS"_2/(color(red)(3)color(red)(cancel(color(black)("moles O"_2)))) = "0.333 moles CS"_2

This means that the reaction will consume $0.333$ moles of carbon disulfide and $1.00$ mole of oxygen gas. The remaining carbon disulfide

$1.00 - 0.333 = 0.67$ moles

will not take part in the reaction, i.e. it will be in excess.