Chemical Equation Balancer

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limiting reagent and percent yield calculator

Limiting reagent, excess reagent and percent yield

Reactants are rarely in the exact ratio of the equation, and whichever runs out first sets how much product forms. The Stoichiometry tab does the math in moles, with CIAAW molar masses, and shows what is left over.

From mass to the reactant that runs out first

The method is always the same: convert each mass to moles (mass divided by molar mass), divide each amount in moles by its coefficient in the balanced equation, and compare. The smallest quotient is the limiting reagent, and that quotient is the extent of reaction, that is, how many "times" the equation happens. Products are the extent multiplied by the coefficients and the molar masses. In N2 + 3H2 → 2NH3, with 28 g of N2 (28/28.014 = 0.9995 mol) and 3 g of H2 (3/2.016 = 1.488 mol), the quotients are 0.9995/1 and 1.488/3 = 0.496. H2 limits, the extent is 0.496 mol and 2 x 0.496 = 0.992 mol of NH3 forms, or 16.9 g.
The excess is what is left of the other reactant: the N2 consumed is 0.496 mol, or 13.9 g, and of the starting 28 g there remain 14.1 g. Lavoisier's check closes: 28 + 3 = 31 g before, and 16.9 g of NH3 plus 14.1 g of leftover N2 also make 31 g. A common trap is thinking equal masses are equal amounts: 10 g of H2 and 10 g of O2 are 4.96 mol and 0.3125 mol. The equation 2H2 + O2 → 2H2O asks for 2 mol of H2 per mole of O2; the quotients are 2.48 (H2) and 0.3125 (O2), so O2 limits, 11.3 g of water forms and 8.7 g of H2 is left over.
Percent yield compares what was obtained in the lab with the theoretical yield, what the calculation predicts: yield = actual mass / theoretical mass x 100. If 12.0 g comes out of a predicted 16.9 g of NH3, the yield is 71%. Yields below 100% are the rule, because of incomplete reactions, purification losses and side reactions; above 100% almost always means a wet or impure product. You can also start from the product: asking for 100 g of iron from Fe2O3 + 3CO → 2Fe + 3CO2, the tool computes how much oxide is needed (143 g), and if you enter only the product it shows the required amount of each reactant.

Cases worked out with the tool

Ammonia: 28 g of N2 and 3 g of H2

Input: N2 + H2 -> NH3 | 28 g N2, 3 g H2
Expected output: limitante H2; 16,9 g de NH3

14.1 g of N2 left; extent 0.496 mol.

Water: 10 g of H2 and 10 g of O2

Input: H2 + O2 -> H2O | 10 g H2, 10 g O2
Expected output: limitante O2; 11,3 g de H2O

8.7 g of H2 left, even though H2 is the lighter one.

Quicklime: 100 g of CaCO3

Input: CaCO3 -> CaO + CO2 | 100 g CaCO3
Expected output: 56,0 g de CaO e 44,0 g de CO2

A single reactant, so it is the limiting one; the sum returns to 100 g.

Iron: 1 kg of Fe2O3 with excess CO

Input: Fe2O3 + CO -> Fe + CO2 | 1000 g Fe2O3
Expected output: 699 g de Fe

1000/159.687 = 6.262 mol of oxide, 2 Fe per Fe2O3.

Percent yield

Input: teórico 16,9 g, real 12,0 g
Expected output: 71%

12.0 / 16.9 x 100 = 71.0%.

Full tool FAQ

How do you balance a chemical equation?

Count the atoms of each element on both sides and adjust the coefficients (never the subscripts) until the counts match. By hand this is done by trial; the tool solves the exact linear system. For Fe + O2 → Fe2O3 the answer is 4Fe + 3O2 → 2Fe2O3, with 4 Fe atoms and 6 O atoms on each side.

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How do I find the limiting reagent?

Divide the moles of each reactant by its coefficient in the balanced equation; the smallest value is the limiting one. Comparing masses or moles without dividing by the coefficient leads to errors whenever the coefficients differ.

Is the excess reagent consumed?

Part of it, yes: the amount that reacts is what the extent of reaction requires. The rest is left at the end. In the ammonia case, the excess N2 uses 13.9 g of the 28 g and leaves 14.1 g.

Can I start from the mass of product I want?

Yes. Fill in only the amount of the product and the tool returns how much of each reactant is needed, in moles and grams. This is what you do to plan the raw material for a synthesis.

Next steps

Atom count, before and after the coefficients
Element	Before (all 1)	After	Check
C	3	1	3	3	✓
H	8	2	8	8	✓
O	2	3	10	10	✓

Atom count, before and after the coefficients

Element

Before (all 1)

After

Check

React.

Prod.

React.

Prod.

✓

From mass to the reactant that runs out first

The method is always the same: convert each mass to moles (mass divided by molar mass), divide each amount in moles by its coefficient in the balanced equation, and compare. The smallest quotient is the limiting reagent, and that quotient is the extent of reaction, that is, how many "times" the equation happens. Products are the extent multiplied by the coefficients and the molar masses. In N2 + 3H2 → 2NH3, with 28 g of N2 (28/28.014 = 0.9995 mol) and 3 g of H2 (3/2.016 = 1.488 mol), the quotients are 0.9995/1 and 1.488/3 = 0.496. H2 limits, the extent is 0.496 mol and 2 x 0.496 = 0.992 mol of NH3 forms, or 16.9 g.

The excess is what is left of the other reactant: the N2 consumed is 0.496 mol, or 13.9 g, and of the starting 28 g there remain 14.1 g. Lavoisier's check closes: 28 + 3 = 31 g before, and 16.9 g of NH3 plus 14.1 g of leftover N2 also make 31 g. A common trap is thinking equal masses are equal amounts: 10 g of H2 and 10 g of O2 are 4.96 mol and 0.3125 mol. The equation 2H2 + O2 → 2H2O asks for 2 mol of H2 per mole of O2; the quotients are 2.48 (H2) and 0.3125 (O2), so O2 limits, 11.3 g of water forms and 8.7 g of H2 is left over.

Percent yield compares what was obtained in the lab with the theoretical yield, what the calculation predicts: yield = actual mass / theoretical mass x 100. If 12.0 g comes out of a predicted 16.9 g of NH3, the yield is 71%. Yields below 100% are the rule, because of incomplete reactions, purification losses and side reactions; above 100% almost always means a wet or impure product. You can also start from the product: asking for 100 g of iron from Fe2O3 + 3CO → 2Fe + 3CO2, the tool computes how much oxide is needed (143 g), and if you enter only the product it shows the required amount of each reactant.

Cases worked out with the tool

Ammonia: 28 g of N2 and 3 g of H2

Input: N2 + H2 -> NH3 | 28 g N2, 3 g H2
Expected output: limitante H2; 16,9 g de NH3

14.1 g of N2 left; extent 0.496 mol.

Water: 10 g of H2 and 10 g of O2

Input: H2 + O2 -> H2O | 10 g H2, 10 g O2
Expected output: limitante O2; 11,3 g de H2O

8.7 g of H2 left, even though H2 is the lighter one.

Quicklime: 100 g of CaCO3

Input: CaCO3 -> CaO + CO2 | 100 g CaCO3
Expected output: 56,0 g de CaO e 44,0 g de CO2

A single reactant, so it is the limiting one; the sum returns to 100 g.

Iron: 1 kg of Fe2O3 with excess CO

Input: Fe2O3 + CO -> Fe + CO2 | 1000 g Fe2O3
Expected output: 699 g de Fe

1000/159.687 = 6.262 mol of oxide, 2 Fe per Fe2O3.

Percent yield

Input: teórico 16,9 g, real 12,0 g
Expected output: 71%

12.0 / 16.9 x 100 = 71.0%.

Full tool FAQ

How do you balance a chemical equation?

Why does an equation need to be balanced?

How does balancing by trial and error work?

How do you balance equations with ions?

How do you balance a redox reaction?

What does "more than one independent solution" mean?

Why does the tool say a substance is on the wrong side or takes no part?

Can I type the equation with coefficients?

How do you find the limiting reagent and the yield?

Does the tool predict the products of a reaction?

Frequently asked questions

How do I find the limiting reagent?

Is the excess reagent consumed?

Part of it, yes: the amount that reacts is what the extent of reaction requires. The rest is left at the end. In the ammonia case, the excess N2 uses 13.9 g of the 28 g and leaves 14.1 g.

Can I start from the mass of product I want?

Yes. Fill in only the amount of the product and the tool returns how much of each reactant is needed, in moles and grams. This is what you do to plan the raw material for a synthesis.