A reaction rearranges atoms and creates none, so every element has the same number on both sides. N2 + 3 H2 → 2 NH3 has 2 nitrogen and 6 hydrogen atoms either side; the tool finds the smallest whole numbers that do it.
In that equation 1 mol of nitrogen (28.01 g) reacts with 3 mol of hydrogen (6.05 g) to make 2 mol of ammonia (34.06 g): the masses are not in the ratio 1 : 3 : 2, though the total is the same on each side.
With 28 g of nitrogen and 8 g of hydrogen the nitrogen runs out first, however much hydrogen is left: 34.0 g of ammonia is the most that can be made, with 1.96 g of hydrogen unused.
Only 46 percent of the mass that goes into the blast furnace reaction Fe2O3 + 3 CO → 2 Fe + 3 CO2 ends up as iron, however good the yield: that is its atom economy, and the rest is carbon dioxide.
The nitrogen is 0.9995 mol and the hydrogen 3.968 mol; dividing by the coefficients, 1 and 3, gives 0.9995 and 1.323, and the smaller, nitrogen, is the limiting reagent. It makes 2 × 0.9995 = 1.9990 mol of ammonia, 34.0 g, and leaves 1.96 g of hydrogen.
Every formula is read into a count of each element, and the balanced coefficients are the whole numbers that make a column of those counts, reactants positive and products negative, add up to nothing for every element at once. The tool solves that exactly with fractions, so the answer is the smallest whole numbers; if the equation cannot be balanced or has more than one solution (two reactions mixed up in one), it says so.
To find the iron in a tonne of rust, Fe2O3 + 3 CO → 2 Fe + 3 CO2: a tonne of Fe2O3 is 1,000,000 g / 159.69 = 6262 mol; the equation gives 2 mol of iron for 1 of the oxide, so 12525 mol of iron, which is 699 kg. Always go through moles: grams to moles, moles by the coefficients, moles to grams.
The reactants are never mixed exactly in proportion, so one runs out first. Divide each one’s moles by its coefficient: the smallest is the limiting reagent, and the others are in excess. It is not the one with the smallest mass: burning 16 g of methane in 48 g of oxygen, the methane is 0.997 mol and the oxygen 1.5 mol, but the oxygen is limiting because 1.5 ÷ 2 is less than 0.997 ÷ 1.
The excess reagent that does not react is left in the flask: here 1.96 g of hydrogen in the ammonia case. The bars show each reactant’s mass in two parts, the part used (green) and the part left (amber), and the mass that goes in equals the mass that comes out (the used reactants equal the products made) because mass is conserved.
The theoretical yield is what the limiting reagent allows if everything goes to plan. The actual yield, what you weigh at the end, is less, because of side reactions, incomplete reaction and losses in handling. If 28 g of nitrogen and 8 g of hydrogen give 27.2 g of ammonia in practice, the percentage yield is 80 percent of the 34.0 g.
The atom economy is the mass of the wanted product over the mass of all the reactants in the balanced equation. Making iron by Fe2O3 + 3 CO → 2 Fe + 3 CO2 has an atom economy of 45.8 percent, because the carbon dioxide, which is not wanted, carries off more than half the mass; making ammonia has 100 percent. A reaction can have a perfect yield and a poor atom economy, which is why chemists try to design reactions with high atom economy.
To make a mass of product at a given yield, divide by the yield to get the theoretical amount, convert to moles, multiply by the coefficients of the reactants over the product’s, and convert back to grams. For 100 g of ammonia at an 80 percent yield the tool aims at 125 g, and gives the masses of nitrogen and hydrogen to weigh out.
Write formulas joined by + with an arrow or an equals sign, such as Ca(OH)2 + H3PO4 -> Ca3(PO4)2 + H2O; brackets, hydrates (a dot) and state symbols such as (g) are accepted and the numbers in front are worked out. Ions and charges are not: a reaction between ions, like Ag+ + Cl−, has to be written with the compounds, AgNO3 + NaCl -> AgCl + NaNO3.
The reaction is taken to go one way, completely, to the products written: no equilibrium (a reversible reaction like ammonia synthesis in practice stops at a mixture), no side reactions and no rate. It does not know which reactions can happen: it will balance any equation whose atoms match, including some that do not occur. Masses use the standard atomic weights to four or five figures, and the tool balances one reaction at a time (an equation hiding two reactions is refused).
Putting numbers in front of the formulas so that every element has the same number of atoms on both sides, because a chemical reaction rearranges atoms and neither makes nor destroys them. The numbers are the smallest whole ones that work, and they are ratios of moles.
Because the coefficients count particles, not grams: 1 mol of nitrogen is 28.01 g and 1 mol of hydrogen 2.02 g, so the equation N2 + 3 H2 → 2 NH3 is about moles. Divide each mass by its molar mass, use the coefficients, and multiply back by the molar mass of what you want.
The reactant that is used up first and so stops the reaction. Divide the moles of each reactant by its coefficient in the balanced equation; the smallest answer is the limiting reagent. The amount of product is set by it, and the others are left over in excess.
The most product the reaction can make, from the amount of the limiting reagent and the balanced equation, if every molecule reacts as written. For 28 g of nitrogen and 8 g of hydrogen it is 34.0 g of ammonia.
Divide the actual yield by the theoretical yield and multiply by 100: if you get 27.2 g of ammonia where 34.0 g is possible, the yield is 80 percent. A yield over 100 percent means the product is impure or wet, or the arithmetic is wrong.
The mass of the wanted product as a percentage of the mass of all the reactants in the balanced equation. It measures how much of what goes in ends up where you want it, however good the yield; the blast furnace reaction for iron has 46 percent, because the rest is carbon dioxide.
Atoms are rearranged, not made or lost, so the total mass of the atoms is the same on both sides; but molecules join or split, so the number of molecules changes: 4 mol of reactants become 2 mol of product in the ammonia reaction, with the same 34.1 g either side.
Every equation whose atoms can be matched can be balanced, but the tool will refuse one that cannot (the atoms are not the same on the two sides, like H2 -> O2) or that is really two reactions at once (for example Mg + O2 + N2 -> MgO + Mg3N2, which has two independent solutions). Balancing says nothing about whether the reaction happens.
Each product has its own coefficient and its own theoretical mass; choose which one you want as the wanted product, and the yield and the atom economy are worked out for it. The other products are made too, in the amounts the equation gives.
Because the limiting reagent has run out and there is nothing left to react with it. The amount left is the amount you started with, less the moles that the equation needed for the limiting reagent’s moles: in the ammonia case 1.96 g of hydrogen.
This page has changed in 1 archived release, but each of those was a site-wide pass, so none is attributable to this tool on its own and none is listed here. That is not a claim that the tool never changed — a release that reworked many pages at once may well have altered this one too. The changelog has them.