Molar Ratio - Getting The Most From Reactants - BBC Bitesize - BBC

In this guide

1. Revise
2. Test

Pages

1. The design stage
2. Molar ratio
3. Moles and solutions
4. Molar volume
5. Percentage yield
6. Atom economy
7. Excess reactants

Molar ratio

A balanced chemical equation can be used to show how the quantities of reactants relate to the quantities of products in the reaction.

Consider the following reaction:

Aluminium can be obtained from aluminium ore (Al2O3) by a process called electrolysis. Oxygen is also produced by this reaction.

The balanced equation for this process is:

\(\mathbf{2}Al_{2}O_{3}\, (aq) \rightarrow \mathbf{4}Al\, (s) + \mathbf{3}O_{2}\, (g)\)

The equation shows you that for every two moles of aluminium oxide oxidised, four moles of aluminium metal are obtained along with three moles of oxygen gas.

The formula mass of each reactant or product can be used to calculate the reacting masses.

Example

Calculate the mass of aluminium formed when 51 g of aluminium oxide are electrolysed.

Firstly, the formula mass of aluminium oxide can be calculated using the formula and masses found in the data book.

This formula triangle can be used to explain the relationship between the formula mass (the mass of one mole) with any mass and the number of moles that it represents.

Using this triangle for the information from the question, we can calculate how many moles of aluminium oxide 51 g is by using the formula

number of moles = mass/formula mass.

The molar ratio from the balanced equation must be considered to tell us how many moles of aluminium will be released.

\(2A{l_2}{O_3} \to 4Al + 3{O_2}\)

\(2\,moles\,A{l_2}{O_3} \to 4\,moles\,Al\)

\(0.5\,moles\,A{l_2}{O_3} \to \frac{{0.5}}{2} \times 4\,moles\,Al\)

\( = 0.25 \times 4\)

\( = 1\,mole\,aluminium\)

To finish off the question, we must change one mole of aluminium into a mass.

So, 55 g of aluminium oxide will produce 27 g of aluminium upon being electrolysed.

Next pageMoles and solutionsPrevious pageThe design stage

More guides on this topic

• Revise: Controlling the rate
• Revise: Chemical energy
• Revise: Equilibria
• Revise: Chemical analysis

Related links

• BBC Science
• BBC News: Science
• SQA Higher Chemistry
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• Royal Society of Chemistry
• Creative Chemistry