Question

A chemical plant uses electrical energy to decompose aqueous solutions of \(\mathrm{NaCl}\) to give \(\mathrm{Cl}_{2}, \mathrm{H}_{2}\), and \(\mathrm{NaOH}\) : $$ 2 \mathrm{NaCl}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{NaOH}(a q)+\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$ 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.

Short answer

The plant produces approximately 42.8 metric tons of \(\mathrm{H}_{2}\) and 1700 metric tons of \(\mathrm{NaOH}\) daily.

Step-by-step solution

Convert \(\mathrm{Cl}_{2}\) from kg to grams.

Given that the plant produces \(1.5 \times 10^{6}\mathrm{~kg}\) of \(\mathrm{Cl}_{2}\), we need to convert this to grams. To do this, we can use the conversion factor \(1\,\mathrm{kg} = 1000\,\mathrm{g}\): $$1.5 \times 10^{6}\,\mathrm{kg} \times \frac{1000\,\mathrm{g}}{1\,\mathrm{kg}} = 1.5 \times 10^{9}\,\mathrm{g}$$ Now we have the amount of \(\mathrm{Cl}_{2}\) in grams: \(1.5 \times 10^{9}\,\mathrm{g}\).

Convert grams of \(\mathrm{Cl}_{2}\) to moles.

Using the molar mass of \(\mathrm{Cl}_{2}\) which is 2(35.45) g/mol = 70.9 g/mol, we will convert the mass of \(\mathrm{Cl}_{2}\) in grams to moles: $$\frac{1.5 \times 10^{9}\,\mathrm{g}}{70.9\,\mathrm{g/mol}} = 2.12 \times 10^{7}\,\mathrm{mol}$$ Now we have the amount of \(\mathrm{Cl}_{2}\) in moles: \(2.12 \times 10^{7}\,\mathrm{mol}\).

Determine the number of moles of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced.

Using the balanced chemical equation, we see that one mole of \(\mathrm{H}_{2}\) and one mole of \(\mathrm{Cl}_{2}\) are produced for every two moles of \(\mathrm{NaOH}\): $$1\,\mathrm{mol~Cl}_{2} \longrightarrow 1\,\mathrm{mol~H}_{2} \longrightarrow 2\,\mathrm{mol~NaOH}$$ Since we know there are \(2.12 \times 10^{7}\,\mathrm{mol}\) of \(\mathrm{Cl}_{2}\) produced, we can use this proportion to find the amounts of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) produced: - \(\mathrm{H}_{2}\): \(2.12 \times 10^{7}\,\mathrm{mol}\) of \(\mathrm{Cl}_{2} \times \frac{1\,\mathrm{mol~H}_{2}}{1\,\mathrm{mol~Cl}_{2}} = 2.12 \times 10^{7}\,\mathrm{mol}\) of \(\mathrm{H}_{2}\) - \(\mathrm{NaOH}\): \(2.12 \times 10^{7}\,\mathrm{mol}\) of \(\mathrm{Cl}_{2} \times \frac{2\,\mathrm{mol~NaOH}}{1\,\mathrm{mol~Cl}_{2}} = 4.24 \times 10^{7}\,\mathrm{mol}\) of \(\mathrm{NaOH}\)

Convert moles of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) to grams.

Using the molar masses of \(\mathrm{H}_{2}\) (2.02 g/mol) and \(\mathrm{NaOH}\) (40.00 g/mol), we will convert the moles of these substances to grams: - \(\mathrm{H}_{2}\): \(2.12 \times 10^{7}\,\mathrm{mol} \times \frac{2.02\,\mathrm{g}}{1\,\mathrm{mol}} = 4.28 \times 10^{7}\,\mathrm{g}\) - \(\mathrm{NaOH}\): \(4.24 \times 10^{7}\,\mathrm{mol} \times \frac{40.00\,\mathrm{g}}{1\,\mathrm{mol}} = 1.7 \times 10^{9}\,\mathrm{g}\)

Convert grams of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) to metric tons.

Finally, we will convert the amounts of \(\mathrm{H}_{2}\) and \(\mathrm{NaOH}\) in grams to metric tons using the conversion factor \(1\,\mathrm{metric~ton} = 1 \times 10^{6}\,\mathrm{g}\): - \(\mathrm{H}_{2}\): \(\frac{4.28 \times 10^{7}\,\mathrm{g}}{1 \times 10^{6}\,\mathrm{g/metric~ton}} = 42.8\,\mathrm{metric~tons}\) - \(\mathrm{NaOH}\): \(\frac{1.7 \times 10^{9}\,\mathrm{g}}{1 \times 10^{6}\,\mathrm{g/metric~ton}} = 1700\,\mathrm{metric~tons}\) So, the plant produces approximately 42.8 metric tons of \(\mathrm{H}_{2}\) and 1700 metric tons of \(\mathrm{NaOH}\) daily.

Key concepts

Mole Concept

The mole concept is a fundamental principle in chemistry that relates the amount of substance to the number of particles, such as molecules, ions, or electrons. One mole, defined by Avogadro's number, is equivalent to approximately \(6.022 \times 10^{23}\) entities. This large number is used because atoms and molecules are incredibly small and cannot be counted directly.

For a better grasp of this concept, imagine the mole as a counting unit similar to a dozen. Just as a dozen refers to 12 items, a mole refers to Avogadro's number of items. The concept enables chemists to work with a manageable unit when dealing with the incredibly large number of atoms or molecules involved in chemical reactions.

Applying this to exercises, you calculate the number of moles by dividing the mass of a substance by its molar mass. In the given exercise, the conversion of mass of \(\mathrm{Cl}_2\) into moles demonstrates this application, serving as the first step in stoichiometric calculations.

Molar Mass

Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It is a bridge between the atomic or molecular scale and the macroscopic, allowing us to measure out amounts of a substance in the laboratory. The molar mass of an element is numerically equal to its relative atomic mass. For a compound, it is the sum of the molar masses of its constituent elements.

In the context of our exercise, understanding molar mass allows us to convert the mass of \( \mathrm{Cl}_2 \) in grams to moles. The molar mass of \( \mathrm{Cl}_2 \) (70.9 g/mol) is twice that of chlorine's atomic mass since \( \mathrm{Cl}_2 \) consists of two chlorine atoms. This calculation is essential as it connects the mass produced by the plant to the concept of moles, setting the stage for further stoichiometric calculations.

Stoichiometric Calculations

Stoichiometric calculations involve quantitatively analyzing the reactants and products of a chemical reaction using the balanced chemical equation. Practically, it provides a way to predict how much of a product will form from a given amount of reactants or vice versa. The stoichiometry of a reaction is the proportion in which substances react according to the balanced chemical equation.

For instance, in the given exercise, the stoichiometry indicated by the balanced chemical equation allows us to determine the amount of \( \mathrm{H}_2 \) and \( \mathrm{NaOH} \) produced when a known mass of \( \mathrm{Cl}_2 \) is formed. By following the molar ratios from the balanced equation, we use the mole concept and molar mass to complete the stoichiometric calculations. The final step involves converting moles back to a more practical unit, such as grams or metric tons, to present our results in a form that is meaningful for the context, in this case, the production quantities of a chemical plant.