Molar ratio
A A chemical equation written using the symbols and formulae of the reactants and products, so that the number of units of each element present is the same on both sides of the arrow. 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 The amount of something per mole. For example, 'molar enthalpy change' is the enthalpy change per mole. 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.
