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?

Take a look at the balanced chemical equation for this reaction

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

As you can see, carbon disulfide, #"CS"_2#, will react with oxygen gas, #"O"_2#, in a #1:color(red)(3)# mole ratio.

This tells you that in order for the reaction to take place, you need to have #color(red)(3)# moles of oxygen gas for every #1# mole of carbon disulfide.

In other words, every mole of carbon disulfide will consume #color(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:color(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.