The mole is the unit for amount of substance. The number of particles in a substance can be found using the Avogadro constant. The mass of product depends upon the mass of limiting reactant.

A reaction finishes when one of the reactants is all used up. The other reactant has nothing left to react with, so some of it is left over:

- the reactant that is all used up is called the limiting reactant
- the reactant that is left over is described as being in excess

The mass of product formed in a reaction depends upon the mass of the limiting reactant. This is because no more product can form when the limiting reactant is all used up.

The maximum mass of product formed in a reaction can be calculated using:

- the balanced equation
- the mass of the limiting reactant, and
- the relative atomic mass (
*A*_{r}) or relative formula mass (*M*_{r}) values of the limiting reactant and the product

12 g of magnesium reacts completely with excess hydrochloric acid to form magnesium chloride and hydrogen:

Mg(s) + 2HCl(aq) → MgCl_{}(aq) + H_{2}(g)

Calculate the maximum mass of hydrogen that can be produced. (*A*_{r} of Mg = 24, *M*_{r} of H_{2} = 2)

amount~of~magnesium = \(\frac{\textup12}{\textup24}\)

= 0.5 mol

Looking at the equation, 1 mol of Mg forms 1 mol of H_{2}, so 0.5 mol of Mg forms 0.5 mol of H_{2}

mass of H_{2} = *M*_{r} × amount

= 2 × 0.5

>= 1 g

- Question
1.0 g of calcium carbonate decomposes to form calcium oxide and carbon dioxide:

CaCO

_{3}(g) → CaO(s) + CO_{2}(g)Calculate the maximum mass of carbon dioxide that can be produced. (

*M*_{r}of CaCO_{3}= 100,*M*_{r}of CO_{2}= 44amount of calcium carbonate =\(\frac{\textup1.0}{\textup100}\)

= 0.01 mol

mass of CO

_{2}= \(M_{r}\times~\textup{amount}\)= 44 × 0.01

= 0.44 g

The stoichiometry of a reaction is the ratio of the amounts of each substance in the balanced equation. It can be deduced or worked out using masses found by experiment.

6.0 g of magnesium reacts with 4.0 g oxygen to produce magnesium oxide, MgO. Deduce the balanced equation for the reaction. (*A*_{r} of Mg = 24, *M*_{r} of O_{2} = 32)

Step | Action | Result | Result |
---|---|---|---|

1 | Write the formulae of the substances | Mg | O_{2} |

2 | Calculate the amounts | \[\frac{6.0}{24}\] = 0.25 mol | \[\frac{4.0}{32}\] = 0.125 mol |

3 | Divide both by the smaller amount | \[\frac{0.25}{0.125}\] = 2 | \[\frac{0.125}{0.125}\] = 1 |

This means that 2 mol of Mg reacts with 2 mol of O_{2}, so the left-hand side of the equation is:

2Mg + O_{2}

Then balancing in the normal way: 2Mg + O_{2} → 2MgO