When an electrical current is passed through a sodium chloride solution, sodium hydroxide can be produced according to the following equation:
\(2\text{NaCl (aq)} + \text{H}_{2}\text{O (l)} → \text{Cl}_{2}\text{(g)} + \text{H}_{2}\text{(g)} + 2\text{NaOH (aq)}\)
What is the maximum mass of sodium hydroxide that can be obtained from \(\text{4,0}\) \(\text{kg}\) of sodium chloride and \(\text{3,0}\) \(\text{kg}\) of water?
Moles of sodium chloride: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{4 000}}{\text{58,4}} \\ & = \text{68,49}\text{ mol} \end{align*}
Moles of water: \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{3 000}}{\text{18}} \\ & = \text{166,67}\text{ mol} \end{align*}
Now we look at the number of moles of product that each reactant can form.
The mole ratio of \(\text{H}_{2}\text{O}\) to \(\text{NaOH}\) is \(1:2\). So the number of moles of \(\text{NaOH}\) that can be produced from water is: \begin{align*} n_{\text{NaOH}} & = n_{\text{H}_{2}\text{O}} \times \frac{\text{stoichiometric coefficient NaOH}}{\text{stoichiometric coefficient H}_{2}\text{O}}\\ & = \text{166,67}\text{ mol} \text{ H}_{2}\text{O} \times \frac{\text{1}\text{ mol} \text{NaOH}}{\text{2}\text{ mol} \text{ H}_{2}\text{O}}\\ & = \text{83,33}\text{ mol} \text{NaOH} \end{align*}
The mole ratio of \(\text{NaCl}\) to \(\text{NaOH}\) is \(2:2\) (or \(1:1\)). So the number of moles of \(\text{NaOH}\) that can be produced from sodium chloride is: \begin{align*} n_{\text{NaOH}} & = n_{\text{NaCl}} \times \frac{\text{stoichiometric coefficient NaOH}}{\text{stoichiometric coefficient NaCl}}\\ & = \text{68,49}\text{ mol} \text{NaCl} \times \frac{\text{2}\text{ mol} \text{NaOH}}{\text{2}\text{ mol} \text{ NaCl}}\\ & = \text{68,49}\text{ mol} \text{NaOH} \end{align*}
Since \(\text{NaCl}\) produces less \(\text{NaOH}\) than is produced from \(\text{H}_{2}\text{O}\), the sodium chloride is the limiting reagent.
We have \(\text{68,49}\) \(\text{mol}\) of \(\text{NaOH}\).
The maximum mass of sodium hydroxide that can be produced is calculated as follows:
\begin{align*} m & = nM\\ & = (\text{68,49})(\text{40}) \\ & = \text{2 739,6}\text{ g}\\ & = \text{2,7396}\text{ kg} \end{align*}
The maximum amount of sodium hydroxide that can be produced is \(\text{2,74}\) \(\text{kg}\).