Limiting reagent problems are very common in chemistry. If you are given certain amounts of reactants for a chemical reaction, chances are that the ratios aren't the needed ones: one of the reactants will run out first and the reaction will stop before the other reactants get used up.

Think of it as making chocolate-chip cookies. You start with flour, butter, eggs, sugar and chocolate chips. If you starting making a large batch but run out of flour after a dozen cookies, you can't make any more no matter how much of the other ingredients you have.

The easiest way to figure out which reactant runs out first is to do the reaction multiple times, assuming each time that it's a different reactant that runs out first. Each reaction will give you a different amount of products: pick the reaction that forms the least products and that is the one that

Example 1: Burn one mole of H₂ and 1 mole of O₂. How much water is formed?

Solution: In any stoichiometry problem, step 1 is to write down the balanced equation:

2H₂ + O₂ -> 2H₂O

Next, assume that the hydrogen runs out first. We start with 1 mole of hydrogen, so we form

1 mole H₂ * (2 moles H₂O/2 moles H₂) = 1 mole H₂O formed

Now let's see what happens if oxygen runs out first. We start with 1 mole of oxygen, so we form

1 mole O₂ * (2 moles H₂O/1 moles O₂) = 2 moles H₂O formed

We form less product if the hydrogen runs out first: it is therefore the limiting reagent and we form only 1 mole of water in this reaction.

Example 2: When antimony and iodine mix, they form antimony(III) iodide. If we mix 3 moles of antimony and 2 moles of iodine, how much antimony(III) iodide do we form?

Solution: First, write down the balanced equation

2Sb + 3I₂ -> 2SbI₃

Next, assume that the antimony runs out first. We start with 3 moles of antimony, so we form

3 moles Sb *(2 moles SbI₃/2 moles Sb) = 3 moles SbI₃

Next, assume that the iodine runs out first. We start with 2 moles of iodine, so we form

2 moles I₂ *(2 moles SbI₃/3 moles I₂) = 4/3 moles SbI₃

We form less product if the iodine runs out first: it is therefore the limiting reagent and we form only 4/3 moles of SbI₃ in this reaction.

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