Question

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?

Answer 1

Oxygen gas, `"O"_2`.

Explanation:

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.