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?

Question

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?

Share with your friends

Talk Doctor Online in Bissoy App

Md Ariful Islam · Answer 1 · Apr 02, 2024 09:48:30

Limiting reagent problems are all about using the 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

$"CS"_text(2(l]) + color(red)(3)"O"_text(2(g]) -> "CO"_text(2(g]) + 2"SO"_text(2(g])$

The $1:color(red)(3)$ mole ratio that exists between carbon disulfide, $"CS"_2$, and oxygen gas tells you that for every $1$ mole of the former, the reaction consumes $color(red)(3)$ moles of the latter.

You can thus distinguish between three possible cases here

if you have less than $color(red)(3)$ moles of oxygen gas for every mole of carbon disulfide, oxygen gas will be the limiting reagent
if you have more than $color(red)(3)$ moles of oxygen gas for every mole of carbon disulfide, carbon disulfide gas will be the limiting reagent
if you have exactly $color(red)(3)$ 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 .

Now, how much carbon disulfide will actually react? Once again, use 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 $"CS"_2$ will react with $18$ moles of $"O"_2$. Notice that you have a $1:1$ mole ratio between $"CS"_2$ and $"CO"_2$. This means that the reaction will produce

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

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?

1 Answers