Limiting reactant calculator with excess left over

The limiting reactant (or limiting reagent) is the reactant that runs out first, so it decides how much product can form. To find it, turn each reactant into moles and divide by its number in the balanced equation: the smallest result is the limiting reactant. This limiting reactant calculator is part of our Stoichiometry Calculator. Type the equation and the amounts in grams or moles, and it balances the equation, names the limiting reactant, gives the theoretical yield and shows how much of everything is left at the end.

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Stoichiometry Calculator

Find reacting masses, limiting reactants, yield and atom economy

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How to find the limiting reactant in the calculator

1. Open the Stoichiometry Calculator. It opens on iron(III) oxide and carbon monoxide; type your own equation in Equation, such as Al + Cl2 -> AlCl3. You don't have to balance it: the calculator does.
2. Under Amounts, type an amount for each reactant and choose grams or moles for each one. Give at least two amounts to find a limiting reactant; a reactant left blank is treated as being in excess.
3. Under Product, choose the product you want. Add the actual mass collected if you want the percentage yield too.
4. Read the result: the Limiting reactant, the Theoretical yield of your product, and the atom economy.
5. Look at the table of moles and masses at the start, the change and the end. The limiting reactant's row is highlighted and ends at 0; the excess reactant's End (g) is how much is left over.

How to find the limiting reactant by hand

1. Balance the equation.
2. Change each reactant's mass to moles: n = m ÷ M.
3. Divide each number of moles by that reactant's coefficient in the equation.
4. The smallest answer is the limiting reactant. The others are in excess.
5. Use the limiting reactant's moles and the mole ratio to find the moles of product, then m = n × M for its mass.

Step 3 is the one people skip. Comparing moles alone, or comparing masses, can pick the wrong reactant, as Example 1 shows.

Worked examples you can check in the calculator

Example 1: equal moles, but one still runs out

5.4 g of aluminium reacts with 14.2 g of chlorine. Type Al + Cl2 -> AlCl3 and the two masses. The calculator balances it as 2Al + 3Cl₂ → 2AlCl₃ and finds the moles:

There are almost exactly the same number of moles of each, but the equation needs 3 Cl₂ for every 2 Al. Dividing by the coefficients gives 0.2001 ÷ 2 = 0.1001 for aluminium and 0.2003 ÷ 3 = 0.06676 for chlorine, so chlorine is the limiting reactant. It makes 0.2003 × 2/3 = 0.1335 mol of AlCl₃, which is 0.1335 × 133.33 = 17.8 g.

The table shows aluminium ending at 0.0666 mol, so 1.80 g of aluminium is left over. The before and after bars are the same length: 19.6 g went in, and 17.8 g of product plus 1.80 g of aluminium came out.

Example 2: a precipitate, with percentage yield

3.40 g of silver nitrate and 1.50 g of sodium chloride are dissolved and mixed, and 2.60 g of silver chloride is filtered off. Type AgNO3 + NaCl -> AgCl + NaNO3, the two masses, choose AgCl as the product and type 2.6 as the actual mass.

Silver nitrate runs out first, so the theoretical yield of AgCl is 0.02001 × 143.32 = 2.869 g, and the percentage yield is 2.60 ÷ 2.869 × 100 = 90.6%. 0.330 g of sodium chloride is left in the solution.

Example 3: the opposite of what the masses suggest

Pick the Burning methane preset: 8 g of methane and 16 g of oxygen, CH₄ + 2O₂ → CO₂ + 2H₂O. There is twice the mass of oxygen, yet oxygen is the limiting reactant: 16 ÷ 31.998 = 0.500 mol, and 0.500 ÷ 2 = 0.250, while methane gives 8 ÷ 16.043 = 0.499. The most carbon dioxide that can form is 0.250 × 44.009 = 11.0 g, and 3.99 g of methane is left unburnt.

Example 4: amounts in moles

0.15 mol of zinc is added to 0.25 mol of hydrochloric acid. Type Zn + HCl -> ZnCl2 + H2, set both units to moles, and choose H2 as the product. Balanced, it is Zn + 2HCl → ZnCl₂ + H₂. Dividing by the coefficients: 0.15 ÷ 1 = 0.15 for zinc and 0.25 ÷ 2 = 0.125 for the acid, so HCl is limiting. It makes 0.125 mol of hydrogen, 0.252 g, and 0.025 mol (1.63 g) of zinc is left.

Theoretical yield and the excess reactant

The theoretical yield is the most product the limiting reactant can make. Once it has run out, the reaction stops, however much of the other reactant is left. To find the excess left over, work out how much of it reacted using the mole ratio, then subtract from what you started with. In Example 1, 0.1335 mol of aluminium reacted, so 0.2001 − 0.1335 = 0.0666 mol remained, and 0.0666 × 26.982 = 1.80 g.

In industry and in the lab, the cheaper reactant is often added in excess on purpose, so that all of the expensive one reacts. For the yield you actually collect, see the percent yield calculator guide; for grams to grams problems with one reactant, see the stoichiometry calculator guide.

What the calculator doesn't do

• Gas volumes and solution concentrations. Amounts go in as grams or moles. Turn a solution into moles first with n = concentration × volume in dm³ (litres), and a gas with n = volume ÷ molar volume (24 dm³ at room temperature and pressure, or 22.4 L at 0 °C and 1 atm), then type the moles.
• Equations it can't balance. Check the formulas are right, with capital letters for each element.
• Equilibrium. It assumes the reaction goes to completion. A reversible reaction never quite uses up its limiting reactant.

Common mistakes

• Comparing masses. The reactant with less mass isn't necessarily limiting. Always change to moles.
• Comparing moles without dividing by the coefficients. In Example 1 the moles were equal, but chlorine still ran out.
• Using the excess reactant for the yield. The product's moles come from the limiting reactant only.
• Using an unbalanced equation. The coefficients are the whole method.
• Forgetting diatomic elements. Chlorine is Cl₂, 70.9 g/mol, not 35.45.

Read next

Chemistry · Moles and quantitiesStoichiometry calculator: limiting reactant and yieldFind the limiting reactant, theoretical yield, percentage yield and atom economy from any equation, balanced for you, with worked examples to check.