# What is the limiting reagent and the moles of CO_2 formed when 11 moles CS_2 reacts with 18 moles O_2 to produce CO_2 gas and SO_2 gas at STP?

## The equation is $C {S}_{2} \left(g\right) + 3 {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) + 2 S {O}_{2} \left(g\right)$?

Feb 14, 2016

${\text{O}}_{2}$ is the limiting reagent and $\text{6 moles}$ of ${\text{CO}}_{2}$ are produced by the reaction.

#### Explanation:

Limiting reagent problems are all about using the mole ratio that exists between the reactants to determine if you have enough of each reactant to allow for their complete consumption.

Take a look at the balanced chemical equation for your reaction

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

The $1 : \textcolor{red}{3}$ mole ratio that exists between carbon disulfide, ${\text{CS}}_{2}$, and oxygen gas tells you that for every $1$ mole of the former, the reaction consumes $\textcolor{red}{3}$ moles of the latter.

You can thus distinguish between three possible cases here

• if you have less than $\textcolor{red}{3}$ moles of oxygen gas for every mole of carbon disulfide, oxygen gas will be the limiting reagent

• if you have more than $\textcolor{red}{3}$ moles of oxygen gas for every mole of carbon disulfide, carbon disulfide gas will be the limiting reagent

• if you have exactly $\textcolor{red}{3}$ moles of oxygen gas for every mole of carbon disulfide, you are not dealing with a limiting reagent*

The problem tells you that $11$ moles of carbon disulfide are allowed to react with $18$ moles of oxygen gas. This many moles of carbon disulfide would require

11 color(red)(cancel(color(black)("moles CS"_2))) * (color(red)(3)" moles O"_2)/(1color(red)(cancel(color(black)("mole CS"_2)))) = "33 moles O"_2

in order to be completely consumed. Since you don't have enough oxygen gas present, you can say that oxygen gas will be the limiting reagent.

Now, how much carbon disulfide will actually react? Once again, use the mole ratio that exists between the two reactants

18 color(red)(cancel(color(black)("moles O"_2))) * "1 mole CS"_2/(color(red)(3)color(red)(cancel(color(black)("moles O"_2)))) = "6 moles CS"_2

So, $6$ moles of ${\text{CS}}_{2}$ will react with $18$ moles of ${\text{O}}_{2}$. Notice that you have a $1 : 1$ mole ratio between ${\text{CS}}_{2}$ and ${\text{CO}}_{2}$. This means that the reaction will produce

6color(red)(cancel(color(black)("moles CS"_2))) * "1 mole CO"_2/(1color(red)(cancel(color(black)("mole CS"_2)))) = color(green)("6 moles CO"_2)