Moles and solutions
The molar ratio of a chemical reaction is also useful for reactions involving solutions.
Calculating the mass or number of moles of a reactant or product can also be achieved by using the following formula triangle.
Example
What mass of sodium sulfate is formed when 250 cm3 of 0.1 mol l-1 sodium hydroxide is neutralised by an excess of sulfuric acid.
\(H_{2}SO_{4} (aq) + 2NaOH (aq)\rightarrow Na_{2}SO_{4} (aq) + 2H_{2}O (l)\)
The question tells you that there is an excess of acid, so the sodium hydroxide will control how much product is obtained. Using the \(n = c \times v\) triangle will give how many moles of sodium hydroxide actually react.
The molar ratio from the balanced equation must be considered to tell us how many moles of sodium sulfate will be formed.
\(H_{2}SO_{4} (aq) + 2NaOH (aq)\rightarrow Na_{2}SO_{4} (aq) + 2H_{2}O (l)\)
\(2\,moles\, NaOH \rightarrow 1\, mole\, Na_{2}SO_{4}\)
\(0.025\,moles\,NaOH \rightarrow 0.025/2 \times 1\,mole\,Na_{2}SO_{4}\)
\( = 0.0125\,moles\,sodium\,sulfate\)
Lastly, the number of moles of sodium sulfate must be converted into a mass.
