# Reactions and moles - Higher

## Limiting reactants

A reaction finishes when one of the 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
• the reactant that is left over is described as being in

The of 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.

## Reacting mass calculations

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

• the
• the mass of the limiting reactant, and
• the Ar () or Mr () values of the limiting reactant and the product

### Worked example

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

Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g)

Calculate the maximum mass of hydrogen that can be produced. (Ar of Mg = 24, Mr of H2 = 2)

Amount of magnesium =

Amount of magnesium =

= 0.5 mol

Looking at the equation, 1 mol of Mg forms 1 mol of H2, so 0.5 mol of Mg forms 0.5 mol of H2.

Mass of H2 = Mr × amount

= 2 × 0.5

= 1 g

Question

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

CaCO3(g) → CaO(s) + CO2(g)

Calculate the maximum mass of carbon dioxide that can be produced. (Mr of CaCO3 = 100, Mr of CO2 = 44)

Amount of calcium carbonate =

= 0.01 mol

Looking at the equation, 1 mol of CaCO3 forms 1 mol of CO2, so 0.01 mol of CaCO3 forms 0.01 mol of CO2.

Mass of CO2 = relative formula mass × amount

= 44 × 0.01

= 0.44 g

## Calculating balancing numbers

The balancing numbers in an equation can be worked out using masses found by experiment.

### Worked example

6.0 g of magnesium reacts with 4.0 g oxygen to produce magnesium oxide, MgO. Deduce the balanced equation for the reaction. (Ar of Mg = 24, Mr of O2 = 32)

StepActionResultResult
1Write the formulae of the substancesMgO2
2Calculate the amounts = 0.25 mol = 0.125 mol
3Divide both by the smaller amount = 2 = 1

This means that 2 mol of Mg reacts with 1 mol of O2, so the left-hand side of the equation is:

2Mg + O2

Then balancing in the normal way gives: 2Mg + O2 → 2MgO

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