Stoichiometry and the limiting reagent

A chemical equation must be balanced: it must have the same number of atoms of each element on both sides, because atoms are neither created nor destroyed. To balance it, change only the coefficients (the numbers in front); never change the subscripts, because that would change the substance itself. For example, phosphorus burns as P₄ + 5O₂ → P₄O₁₀.

A balanced equation can be read in several ways. Take 2H₂ + O₂ → 2H₂O:

• 2 mol of hydrogen react with 1 mol of oxygen to give 2 mol of water.
• In mass: 4 g of hydrogen react with 32 g of oxygen to give 36 g of water.
• In volume (for gases at STP, where one mole of gas takes up 22.7 L): 45.4 L of hydrogen react with 22.7 L of oxygen.

Stoichiometry is calculating the amounts of substances in a reaction. The method is always the same: change the given amount into moles, use the mole ratio from the balanced equation, then change the answer back into grams or litres. Example: how much water forms from 8 g of hydrogen? 8 g ÷ 2 g mol⁻¹ = 4 mol H₂; the equation gives 2 mol water for 2 mol H₂, so 4 mol water = 4 × 18 = 72 g.

Often one reactant runs out before the other. The limiting reagent is the reactant that is used up first; it decides how much product can form. The other is in excess, and some is left over. It is like making sandwiches: with 10 slices of bread and 8 slices of cheese, two slices of bread per sandwich, you can make only 5 sandwiches. The bread limits you, and 3 slices of cheese are left over.

To find the limiting reagent, change both amounts into moles, work out how much of one reactant the other one needs, and compare that with what is actually there.