Chemical Equation Balancer
Type the unbalanced equation into the chemical equation balancer. The coefficients come from exact linear algebra, every element is checked on both sides, and the reactant masses below turn it into a yield calculation.
Balanced equation
- Balanced equation The smallest whole-number coefficients that conserve every element.
- 4Fe + 3O2 → 2Fe2O3
- Coefficients One row per element (2 equations) in 3 unknowns, solved exactly. The coefficients are mole ratios: 4 mol of Fe to 3 mol of O2, and so on.
- Fe: 4, O2: 3, Fe2O3: 2
- Fe: 10 g 10.00 g ÷ 55.845 g/mol. Divided by its coefficient, 4: 0.04477. That is the smallest, so this reactant runs out first.
- 0.1791 mol
- O2: 5 g 5.000 g ÷ 31.998 g/mol. Divided by its coefficient, 3: 0.05209.
- 0.1563 mol
- Limiting reactant The limiting reactant is the one with the fewest moles per unit of its coefficient, not the one with the smallest mass. Fe gives 0.04477 against 0.05209 for O2, so it runs out first and everything else is left in excess.
- Fe
- Theoretical yield of Fe2O3 0.1791 mol of Fe × 2/4 = 0.08953 mol of Fe2O3, × 159.687 g/mol = 14.30 g. The most the reaction can make if it goes to completion with no losses.
- 14.30 g
- O2 left over 0.1343 mol of the 0.1563 mol is consumed; the rest remains when Fe runs out.
- 0.7027 g
- Percent yield Actual ÷ theoretical × 100 = 12.00 g ÷ 14.30 g × 100. Losses in transfer, filtration and side reactions are why it is below 100%.
- 83.93%
- Molar masses (g/mol) From the IUPAC (CIAAW 2024) abridged standard atomic weights.
- Fe 55.845, O2 31.998, Fe2O3 159.687
| Element | Reactants | Products | Balanced |
|---|---|---|---|
| Fe | 4 | 4 | yes |
| O | 6 | 6 | yes |
- Balancing conserves atoms and charge. It does not say whether the reaction happens, which products form, or how far it goes.
How the Chemical Equation Balancer works
A balanced equation has the same number of atoms of every element on each side, because a chemical reaction rearranges atoms and does not create or destroy them. Balancing means finding the whole-number coefficients that make that true. For iron rusting, Fe + O2 → Fe2O3: let the coefficients be a, b and c. Iron gives a = 2c, oxygen gives 2b = 3c. Choose c = 2, and a = 4, b = 3: 4Fe + 3O2 → 2Fe2O3, with four iron and six oxygen atoms on each side.
That is the method this balancer uses, done systematically. Each element is one equation and each species is one unknown, so the equation becomes a matrix with a row per element and a column per species, products counted negative. The coefficients are the solution of that matrix times the coefficient vector equals zero (its nullspace), found by Gauss-Jordan elimination in exact fractions and scaled to the smallest whole numbers. Exact arithmetic matters: in decimals, 2/3 becomes 0.6666667 and has to be guessed back. Balancing by inspection, the method taught first, finds the same answer on simple equations and gets stuck on the ones with five or six species, such as KMnO4 + HCl → KCl + MnCl2 + H2O + Cl2, which is 2, 16, 2, 2, 8, 5.
Ionic equations add one more rule: charge is conserved too. Write charges as Fe^3+ or SO4^2-, or as Fe3+ with spaces around the plus signs that separate species, and electrons as e-. The total charge becomes one more row of the matrix. For a redox reaction in acid, include H+ and H2O yourself: MnO4^- + Fe^2+ + H^+ → Mn^2+ + Fe^3+ + H2O balances to 1, 5, 8, 1, 5, 4.
Some equations do not have one answer. If no positive coefficients work, an element appears on one side only or a formula is mistyped, and the page says which. If several independent balances exist (H2 + O2 → H2O + H2O2 can make any mix of water and peroxide), the equation is underdetermined: it is really two reactions written as one. The page shows the balance with the smallest coefficients and says so.
The coefficients are mole ratios, and that is what turns a balanced equation into a yield calculation. Convert each reactant's mass to moles by dividing by its molar mass, then divide by its coefficient. The reactant with the smallest result is the limiting reactant: it runs out first and caps how much product can form. In the example, 10 g of iron is 0.1791 mol, or 0.04477 per unit of its coefficient of 4; 5 g of oxygen is 0.1563 mol, or 0.05209 per unit of 3. Iron runs out first.
Theoretical yield is the product the limiting reactant can make: 0.1791 mol of Fe × 2/4 = 0.08953 mol of Fe2O3, × 159.69 g/mol = 14.30 g. Percent yield compares what you collected with that ceiling: percent yield = actual yield ÷ theoretical yield × 100, so 12 g collected is 83.93%. A yield above 100% is a sign the product is wet or impure rather than a better-than-perfect reaction.
The limiting reactant is not the one with the smallest mass. It is the one with the fewest moles relative to what the equation asks for, which is why the masses have to go through molar mass and the coefficients before they can be compared. Molar masses here are summed from the IUPAC standard atomic weights.
Chemical Equation Balancer FAQ
How do you balance a chemical equation?
Give each species an unknown coefficient and write one equation per element: atoms on the left equal atoms on the right. Solve for the smallest whole numbers. Fe + O2 → Fe2O3 gives a = 2c and 2b = 3c, so 4Fe + 3O2 → 2Fe2O3. This balancer does the same thing with exact matrix arithmetic, which also handles equations too large to balance by inspection.
How do you calculate theoretical yield?
Find the limiting reactant, convert its mass to moles, multiply by the product's coefficient over the reactant's coefficient, and multiply by the product's molar mass. 10 g of iron (0.1791 mol) gives 0.1791 × 2/4 = 0.08953 mol of Fe2O3, which is 14.30 g.
What is the percent yield formula?
Percent yield = actual yield ÷ theoretical yield × 100. If the reaction above produced 12 g of Fe2O3 against a theoretical 14.30 g, the percent yield is 83.93%. Both yields must be in the same unit.
How do I find the limiting reactant?
Divide the moles of each reactant by its coefficient in the balanced equation. The smallest result is the limiting reactant. It is not necessarily the reactant with the smallest mass: 10 g of iron limits against only 5 g of oxygen because iron's molar mass is higher and the equation needs four irons for every three O2.
Can every equation be balanced?
No. If an element appears on only one side, or a formula is mistyped, no set of positive coefficients works and the page says why. Some equations have several independent balances; those are usually two reactions written as one, and each should be balanced on its own.