Reacting masses and limiting reactants

The numbers in a balanced equation give the mole ratio of reactants and products. Use it to work out masses:

1. Find the moles of the substance you know.
2. Use the ratio from the equation to find the moles of the one you want.
3. Convert to mass (mass = moles × Mᵣ).

Example: what mass of magnesium oxide is made from 12 g of magnesium?

2Mg + O₂ → 2MgO (Mg = 24, MgO = 40)

1. moles of Mg = 12 ÷ 24 = 0.5 mol
2. ratio Mg : MgO = 2 : 2, so 0.5 mol of MgO
3. mass = 0.5 × 40 = 20 g

A reaction stops when one reactant is used up. That is the limiting reactant: it decides how much product forms. The other reactant is in excess — some is left over.

Example: 4.8 g of Mg reacts with 7.3 g of HCl. What mass of hydrogen forms?

Mg + 2HCl → MgCl₂ + H₂ (Mg = 24, HCl = 36.5, H₂ = 2)

1. moles of Mg = 4.8 ÷ 24 = 0.2 mol; moles of HCl = 7.3 ÷ 36.5 = 0.2 mol
2. 0.2 mol of Mg needs 0.4 mol of HCl, but there's only 0.2 mol, so HCl is limiting.
3. 2 mol of HCl gives 1 mol of H₂, so 0.2 mol gives 0.1 mol of H₂.
4. mass = 0.1 × 2 = 0.2 g

If you know the masses that react, you can work out the balancing numbers:

1. Convert each mass to moles.
2. Divide each by the smallest number of moles.
3. Use these whole numbers to balance the equation.

Example: 4.8 g of Mg reacts with 3.2 g of O₂ to make 8.0 g of MgO (Mg = 24, O₂ = 32, MgO = 40).

1. Mg = 4.8 ÷ 24 = 0.2 mol; O₂ = 3.2 ÷ 32 = 0.1 mol; MgO = 8.0 ÷ 40 = 0.2 mol
2. ÷ 0.1 → Mg 2, O₂ 1, MgO 2
3. 2Mg + O₂ → 2MgO