Stoichiometry step by step, with limiting reactants

Almost every stoichiometry problem is the same three conversions: grams of what you have to moles, moles of what you have to moles of what you want, and moles back to grams. The balanced equation supplies the middle step. Once that path is automatic, limiting reactant and percent yield are small additions to it.

All the examples below use the ammonia synthesis reaction:

N2(g) + 3 H2(g) -> 2 NH3(g)

and these molar masses (IUPAC atomic weights, the same ones the molar mass calculator uses): N2 = 28.014 g/mol, H2 = 2.016 g/mol, NH3 = 17.031 g/mol. If your course's periodic table rounds to H 1.01, your answers may differ in the last digit.

Step 0: balance the equation

Mole ratios come only from a balanced equation. Balance by changing coefficients, never subscripts. Here the equation says 1 mol N2 reacts with 3 mol H2 to make 2 mol NH3. Those coefficients are mole ratios, not mass ratios: 1 g of N2 does not react with 3 g of H2.

Step 1: grams to moles

Divide the mass by the molar mass.

moles = mass (g) / molar mass (g/mol)

For example, 36.0 g of water (18.015 g/mol) is 36.0 / 18.015 = 2.00 mol.

Step 2: moles to moles with the mole ratio

Multiply by the ratio of coefficients, with what you want on top and what you have on the bottom.

Step 3: moles to grams

Multiply by the molar mass of the product.

Example 1: how much ammonia from 50.0 g of nitrogen?

Assume plenty of H2.

Moles of N2: 50.0 g / 28.014 g/mol = 1.785 mol
Moles of NH3: 1.785 mol N2 x (2 mol NH3 / 1 mol N2) = 3.570 mol
Grams of NH3: 3.570 mol x 17.031 g/mol = 60.8 g NH3

Written as one line of dimensional analysis:

50.0 g N2 x (1 mol N2 / 28.014 g N2) x (2 mol NH3 / 1 mol N2) x (17.031 g NH3 / 1 mol NH3) = 60.8 g NH3

Setting it up this way lets the units cancel, which catches an upside-down ratio before you finish.

Limiting reactant: what runs out first

When a problem gives amounts of two reactants, one of them usually runs out first. That one is the limiting reactant, and it decides how much product can form. The other is in excess, and some of it is left over.

The reliable method: work out how much product each reactant could make on its own. Whichever gives less product is limiting.

Don't compare the grams of the reactants directly, and don't compare their moles without the mole ratio. Hydrogen is light, so a few grams of it is a lot of moles.

Example 2: 25.0 g N2 and 5.00 g H2

Moles of each reactant:

N2: 25.0 / 28.014 = 0.8924 mol
H2: 5.00 / 2.016 = 2.480 mol

Product each could make:

From N2: 0.8924 x (2 / 1) = 1.785 mol NH3
From H2: 2.480 x (2 / 3) = 1.653 mol NH3

H2 gives less, so H2 is the limiting reactant, even though it has more moles than N2. The 1:3 ratio means N2 needs three times its own moles in hydrogen, and 3 x 0.8924 = 2.677 mol is more H2 than you have.

Theoretical yield of NH3:

1.653 mol x 17.031 g/mol = 28.2 g NH3

Example 3: how much excess reactant is left?

Use the limiting reactant to find how much of the excess reactant actually reacts.

N2 used: 2.480 mol H2 x (1 mol N2 / 3 mol H2) = 0.8267 mol N2, which is 0.8267 x 28.014 = 23.16 g
N2 left: 25.0 - 23.16 = 1.84 g N2

Check with conservation of mass: 23.16 g of N2 plus 5.00 g of H2 reacted, which is 28.16 g, matching the 28.2 g of NH3 formed. The other 1.84 g of N2 is left over. The totals agree, so the setup is right.

Example 4: percent yield

Real reactions rarely make the full theoretical yield. Some product is lost, side reactions happen, or the reaction doesn't go to completion.

percent yield = (actual yield / theoretical yield) x 100%

If the reaction in Example 2 actually produced 24.0 g of NH3:

24.0 / 28.16 x 100% = 85.2%

Percent yield uses the theoretical yield from the limiting reactant, never from the excess one. A percent yield over 100% in a lab usually means the product was still wet or impure.

Where students lose points

Skipping the balance. An unbalanced equation gives wrong mole ratios, and every later number is wrong with it.
Flipping the mole ratio. Put what you want on top. If the units don't cancel, it's upside down.
Comparing grams to find the limiting reactant. Compare the product each reactant can make.
Using the excess reactant for the theoretical yield. Only the limiting reactant sets the maximum.
Rounding every step. Keep extra digits and round the final answer to the right number of significant figures. Three significant figures in the data means three in the answer.
Using the wrong molar mass for diatomic elements. N2 is 28.014 g/mol, not 14.007. Hydrogen, nitrogen, oxygen, fluorine, chlorine, bromine and iodine are diatomic as elements.

Solutions and gases use the same path

Stoichiometry doesn't always start in grams. For a solution, moles = molarity x volume in liters. For a gas, moles come from PV = nRT. The middle step, the mole ratio, never changes. See C1V1 = C2V2 explained for concentration math and the ideal gas law with worked examples for gases. Once equilibrium shows up in the second semester, ICE tables step by step build on the same mole bookkeeping. Biology uses it too: the overall equation for cellular respiration, C6H12O6 + 6 O2 -> 6 CO2 + 6 H2O, is a balanced equation you can run mole ratios on.

Practice until the path is automatic

Stoichiometry rewards repetition more than insight. Do several problems a day for a week instead of twenty the night before, which is the point of spaced repetition. Check your molar masses with the molar mass calculator so you're practicing the method, not hunting for arithmetic slips.

Encodr's free General Chemistry I flashcards have full units on the mole and on stoichiometry, including limiting reactant, yield and titration problems.