Limiting Reactants - the Math

Limiting Reactant problems have one primary difference from other stoichiometry problems - they present amounts of more than one reactant. This is how we can immediately recognize them. Here’s an example:

When 13.9 g of copper II nitrate react with 34.9 mL of 1.07M potassium phosphate, how many molecules of copper II phosphate are created?

Let’s start, as we always do, by writing the balanced reaction and putting the relevant information beneath the compounds.

Because we have two starting amounts, we know that this is a limiting reactant problem. 

Solving a limiting reactant problem looks almost the same as a regular stoichiometry problem with two straightforward differences: 

1) We have to solve more than one stoichiometry problem (one problem for each given amount) 

2) We have to pick which answer is correct (It’s ALWAYS the smaller one) 

So, let’s go to work. First we'll solve the two problems:

So, we have two answers to the question: 1.49x10²² mc and 1.12x10²² mc of Cu₃(PO₄)₂. We have to pick one, but that choice is easy. As stated above, the correct answer is ALWAYS the smaller one. 

So, the answer is 1.12x10²² mc of Cu₃(PO₄)₂.

One last thought. Since the K₃PO₄ limited our reaction (gave us the smaller answer), it is called the limiting reactant. Note, the the limiting reactant is NOT the smaller answer, it is the reactant that led to the smaller answer.