A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NaOH}\) : $$ \begin{aligned} \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow & \\ & 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \end{aligned} $$ If the plant produces \(1.5 \times 10^{6} \mathrm{~kg}(1500\) metric tons \()\) of \(\mathrm{Cl}_{2}\) daily, estimate the quantities of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.

In order to solve this problem, we need to have a balanced chemical equation. The given equation is already balanced, as shown below: \(2 NaCl(aq) + 2 H_2O(l) \longrightarrow 2 NaOH(aq) + H_2(g) + Cl_2(g)\)

We are given the mass of Cl₂ produced (1.5 x 10⁶ kg). To make use of stoichiometry, we need to convert the mass into moles. We can do that using the molar mass of Cl₂, which is 35.45 x 2 = 70.9 g/mol: moles of Cl₂ = \(\frac{mass}{molar~mass}\) = \(\frac{1.5\times10^{6}~kg}{70.9~g/mol}\) As we're dealing with units involving kg and g: moles of Cl₂ = \(\frac{1.5\times10^{6}~kg \times 10^{3}~g/kg}{70.9~g/mol}\) moles of Cl₂ = \(2.118 \times 10^{7}~mol\)

Using stoichiometry and the balanced chemical equation, we can determine the moles of H₂ produced by comparing the stoichiometric coefficients: \(2 NaCl(aq) + 2 H_2O(l) \longrightarrow 2 NaOH(aq) + H_2(g) + Cl_2(g)\) From the balanced equation, we can observe that 1 mole of H₂ is produced for every 1 mole of Cl₂. Therefore, the moles of H₂ produced is equal to the moles of Cl₂: moles of H₂ = moles of Cl₂ = \(2.118 \times 10^{7}~mol\)

Similarly, we can determine the moles of NaOH produced using stoichiometry. From the balanced equation, we can observe that 2 moles of NaOH are produced for every 1 mole of Cl₂. Therefore, we can calculate the moles of NaOH produced: moles of NaOH = 2 x moles of Cl₂ = 2 x \(2.118 \times 10^{7}~mol\) moles of NaOH = \(4.236 \times 10^{7}~mol\)

Finally, to obtain the quantities of H₂ and NaOH produced, we need to convert the moles back into mass using their respective molar masses. The molar mass of H₂ is 2.016 g/mol, and that of NaOH is 22.99 + 15.999 + 1.008 = 39.998 g/mol. So, mass of H₂ produced = moles of H₂ x molar mass of H₂ mass of H₂ produced = \(2.118 \times 10^{7}~mol \times 2.016~g/mol\) mass of H₂ produced = \(4.269 \times 10^{7}~g\) or \(4.269 \times 10^{4}~kg\) mass of NaOH produced = moles of NaOH x molar mass of NaOH mass of NaOH produced = \(4.236 \times 10^{7}~mol \times 39.998~g/mol\) mass of NaOH produced = \(1.694 \times 10^{9}~g\) or \(1.694 \times 10^{6}~kg\) So, the estimated quantities of H₂ and NaOH produced are approximately 42,690 kg of H₂ and 1,694,000 kg of NaOH.