When an electrical current is passed through a sodium chloride solution, sodium hydroxide can be produced according to the following equation:

\[2\text{NaCl(aq)} + 2\text{H}_{2}\text{O(ℓ)} → \text{Cl}_{2}\text{(g)} + \text{H}_{2}\text{(g)} + 2\text{NaOH(aq)}\]

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*} M_{\text{NaCl}} &= \text{23} + \text{35,45} \\ &= \text{58,45 g·mol$^{-1}$} \end{align*} \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{4 000}}{\text{58,45}} \\ & = \text{68,434}\ldots\text{ mol} \end{align*}

Moles of water: \begin{align*} M_{\text{H}_{2}\text{O}} &= \text{1,01} \times \text{2} + \text{16} \\ &= \text{18,02 g·mol$^{-1}$} \end{align*} \begin{align*} n & = \frac{m}{M}\\ & = \frac{\text{3 000}}{\text{18,02}} \\ & = \text{166,481}\ldots\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 \(2:2\). So the number of moles of \(\text{NaOH}\) that can be produced from the available 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{166,481}\ldots\text{ mol} \text{ H}_{2}\text{O} \times \frac{\text{2}\text{ mol } \text{NaOH}}{\text{2}\text{ mol } \text{ H}_{2}\text{O}}\\ & = \text{166,481}\ldots\text{ mol } \text{NaOH} \end{align*}

The mole ratio of \(\text{NaCl}\) to \(\text{NaOH}\) is also \(2:2\) (or \(1:1\)). So the number of moles of \(\text{NaOH}\) that can be produced from the available sodium chloride is: \begin{align*} n_{\text{NaOH}} & = n_{\text{NaCl}} \times \frac{ \text{stoichiometric coefficient NaOH}}{\text{stoichiometric coefficient NaCl}}\\ & = \text{68,434}\ldots\text{ mol } \text{NaCl} \times \frac{\text{2}\text{ mol } \text{NaOH}}{\text{2}\text{ mol } \text{ NaCl}}\\ & = \text{68,434}\ldots\text{ mol } \text{NaOH} \end{align*}

Since the available \(\text{NaCl}\) can produce less \(\text{NaOH}\) than the available \(\text{H}_{2}\text{O}\), the sodium chloride is the limiting reagent.

We have \(\text{68,434}\ldots\) \(\text{mol}\) of \(\text{NaOH}\).

The maximum mass of sodium hydroxide that can be produced is calculated as follows:

\begin{align*} M_{\text{NaOH}} &= \text{23} + \text{16} + \text{1,01} \\ &= \text{40,01 g·mol$^{-1}$} \end{align*}\begin{align*} m & = nM\\ & = (\text{68,434}\ldots)(\text{40,01}) \\ & = \text{2 738,066}\ldots\text{ g}\\ & = \text{2,7380}\ldots\text{ kg} \end{align*}

The maximum amount of sodium hydroxide that can be produced is \(\text{2,74}\) \(\text{kg}\).