Since first writing this webpage I've modified how I teach limiting reactants even more. It's the same method as outlined further below but more straightforward. Here's how to do it:

1) Divide the moles of each reactant by its stoichiometric coefficient from the balanced chemical equation. Technically, this converts moles of reactant to moles of reaction (what I label x below).

2) The reactant that results in the smallest number of moles of reaction, x, from step #1 is the limiting reactant.

3) In order to calculate how much of the other reactants is used during the reaction, just multiply x by the stoichiometric coefficient of the reactant you're interested in. The difference between the amount of reactant you start with and how much gets used is the excess of that reactant.

4) In order to calculate how much of a product is made during a reaction, just multiply x by the stoichiometric coefficient of the product you're interested in.

The primary difference between the procedure outlined above and what I originally did below is the focus below is on calculating the amount of excess reactant while the procedure above calculates how much reactant is used.
 Reaction Initial Limiting Stoichiometry Final

We will use the synthesis of ammonia, NH3, from nitrogen gas, N2, and hydrogen gas, H2, as an example of the method.

The balanced reaction goes in the Reaction row.

The initial amounts, generally in moles, go in the Initial row. I have just made up some numbers.

The Limiting Stoichiometry row subtracts stoichiometric amounts of "x" from the reactant species and adds stoichiometric amounts of "x" for product species.

The Final row is simply the sum of the Initial row and the Limiting Stoichiometry row.

It is very important to understand what "x" means.

"x" is the number of times the reaction happens, given in the same units being used to count the reactants and products (probably moles).
 Reaction 3 H2 (g) + N2 (g) --> 2 NH3 (g) Initial 6 mol 4 mol 0 mol Limiting Stoichiometry -3x -x +2x Final 6 - 3x 4 - x 2x

For example, for every reaction that happens, 2 NH3 will be made, so the number of NH3 ultimately made is equal to 2 times the number of times the reaction happens, 2x.

The only thing to do now is figure out what "x" is equal to.

This is done by dividing the moles of each reactant by its own stoichiometric coefficient from the balanced chemical equation and then picking the smallest value of x obtained from all the different reactants. You have to pick the smallest value of x because that x came from the reactant that you will run out of first, thus limiting how many times your reaction can happen.

The reactant species that gives you the smallest value for x using this procedure is the Limiting Reactant.

(6-3x) = 0 and (4-x) = 0

x = 6/3 = 2 or x = 4

x is thus equal to 2 for our reaction, making H2 the Limiting Reactant. Thus, N2 is an Excess Reactant.

It is now trivial to calculate all the things that these sort of questions generally ask you.

How much NH3 is made? Just grab its entry in the Final row, 2x, and do the math: 2(2) = 4 mol NH3 is produced

How much H2 is left over? Well, that's easy, 0 mol, because it was the limiting reactant. But the math shows us this as well: 6 - 3(2) = 6 - 6 = 0 mol H2 is left over

How much N2 is left over? 4 - 2 = 2 mol N2 is left over

Of course, this assumes the reaction proceeds to completion, ignoring the concept of dynamic equilibrium entirely. Though it's not a terrible approximation for those reactions with enormous equilibrium constants.

For those that would use this as a precursor to teaching how to solve problems using the RICE method, let's assume the volume is 1 L so all values become molarities.

Keq = ( [NH3]eq2 ) / ( [H2]eq3 [N2]eq ) = (2x)2 / (6-3x)3 (4-x) = 5.72 * 105 ~ infinity

The first step to solve this problem using the limiting reactant algorithm is to just take the denominator and set it equal to zero, since dividing by zero is the fastest way to get to infinity.

(6-3x)^3 (4-x) = 0

Any exponents don't matter at all.

(6-3x) (4-x) = 0

There are two solutions to this equation, having set each parenthetical equation equal to the 0 on the right hand side.

(6-3x) = 0 and (4-x)=0

The rest is the same as above, except we'll be in molarity instead of moles due to assuming the volume of the container this is being carried out in is 1 L.

You can compare these values to the true equilibrium concentrations where x = 1.992.

[H2]eq = 6 - 3x = 6 - 3(1.992) = 0.024 M H2

[N2]eq = 4 - x = 4 - 1.992 = 2.008 M N2

[NH3]eq = 2x = 2(1.992) = 3.984 M NH3

For such large equilibrium constants, the limiting reactant approximation isn't terrible.