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?

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

#"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 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 #"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)#