91Ó°ÊÓ

The balanced chemical equation for the decomposition of aqueous solutions of NaCl is given as: $$ 2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$

We are given that the plant produces \(1.5 \times 10^{6} \mathrm{kg}\) of Clâ‚‚ daily. To find the moles of Clâ‚‚, we can use the following formula: $$ \text{moles of Clâ‚‚} = \frac{\text{mass of Clâ‚‚}}{\text{molar mass of Clâ‚‚}} $$ The molar mass of Clâ‚‚ is \(2 \times 35.45 \thinspace \mathrm{g/mol}\) (since there are two chlorine atoms, each with a molar mass of \(35.45 \thinspace \mathrm{g/mol}\)). First, convert the mass of Clâ‚‚ from kg to g: $$ 1.5 \times 10^{6} \thinspace \mathrm{kg} \times \frac{1000 \thinspace \mathrm{g}}{1 \thinspace \mathrm{kg}} = 1.5 \times 10^{9} \thinspace \mathrm{g} $$ Next, calculate the moles of Clâ‚‚ produced daily: $$ \text{moles of Clâ‚‚} = \frac{1.5 \times 10^{9} \thinspace \mathrm{g}}{2 \times 35.45 \thinspace \mathrm{g/mol}} = \frac{1.5 \times 10^{9} \thinspace \mathrm{g}}{70.9 \thinspace \mathrm{g/mol}} = 2.12 \times 10^{7} \thinspace \mathrm{mol} $$

From the balanced equation, we can see that for each mole of Clâ‚‚ produced, one mole of Hâ‚‚ is produced, and two moles of NaOH are produced. Therefore: Moles of Hâ‚‚ produced daily = Moles of Clâ‚‚ produced daily = \(2.12 \times 10^{7} \thinspace \mathrm{mol}\) Moles of NaOH produced daily = \(2 \times\) Moles of Clâ‚‚ produced daily = \(2 \times 2.12 \times 10^{7} \thinspace \mathrm{mol} = 4.24 \times 10^{7} \thinspace \mathrm{mol}\)

Now we can calculate the mass of Hâ‚‚ and NaOH produced daily using their respective molar masses: Molar mass of Hâ‚‚ = \(2 \times 1.01 \thinspace \mathrm{g/mol}\) (since there are two hydrogen atoms, each with a molar mass of \(1.01 \thinspace \mathrm{g/mol}\)) Molar mass of NaOH = \(22.99 \thinspace \mathrm{g/mol} + 15.999 \thinspace \mathrm{g/mol} + 1.01 \thinspace \mathrm{g/mol}\) (Na, O, and H) Mass of Hâ‚‚ produced daily: $$ (2.12 \times 10^{7} \thinspace \mathrm{mol}) \times (2.02 \thinspace \mathrm{g/mol}) = 4.28 \times 10^{7} \thinspace \mathrm{g} = 4.28 \times 10^{4} \thinspace \mathrm{kg} $$ Mass of NaOH produced daily: $$ (4.24 \times 10^{7} \thinspace \mathrm{mol}) \times (39.998\thinspace \mathrm{g/mol}) = 1.69 \times 10^{9} \thinspace \mathrm{g} = 1.69 \times 10^{6} \thinspace \mathrm{kg} $$

The chemical plant produces approximately \(4.28 \times 10^{4} \thinspace \mathrm{kg}\) of hydrogen gas (Hâ‚‚) and \(1.69 \times 10^{6} \thinspace \mathrm{kg}\) of sodium hydroxide (NaOH) daily, along with the given \(1.5 \times 10^6 \thinspace \mathrm{kg}\) of chlorine gas (Clâ‚‚).