Problem 100 - Chemistry: The Central Science in SI Units, Global Edition

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.

Step-by-Step Solution

Answer

In conclusion, using stoichiometric calculations, we found that the chemical plant produces daily approximately 2.14 x 10⁴ kg (21.4 metric tons) of H₂ and 8.48 x 10⁵ kg (848 metric tons) of NaOH.

Step 1: Calculate moles of Cl₂ produced daily

First, we must convert the mass of Cl₂ produced daily (1.5 x 10⁶ kg) into moles, using the molar mass of Cl₂ (70.90 g/mol). Remember, we have to convert from kg to grams before the conversion: 1.5 x 10⁶ kg * 1000 g/kg = 1.5 x 10⁹ g. Now, we can find the moles of Cl₂ produced: \[\frac{1.5 \times 10^{9} \mathrm{~g} \mathrm{Cl}_{2}}{70.90 \mathrm{~g/mol}} \approx 2.12 \times 10^{7} \mathrm{~moles} \mathrm{Cl}_{2}\]

Step 2: Determine moles of H₂ produced daily

From the balanced chemical equation, we can see that one mole of H₂ is produced for every two moles of Cl₂: 2 NaCl + 2 H₂O → 2 NaOH + H₂ + Cl₂. So, the moles of H₂ produced daily are half the amount of moles of Cl₂ produced daily: \[2.12 \times 10^{7} \mathrm{~moles} \mathrm{Cl}_{2} \times \frac{1 \mathrm{~mole} \mathrm{H}_{2}}{2 \mathrm{~moles} \mathrm{Cl}_{2}} = 1.06 \times 10^{7} \mathrm{~moles} \mathrm{H}_{2}\]

Step 3: Calculate the mass of H₂ produced daily

Now, we need to convert the moles of H₂ to mass, using the molar mass of H₂ (2.02 g/mol). \[1.06 \times 10^{7} \mathrm{~moles} \mathrm{H}_{2} \times \frac{2.02 \mathrm{~g/mol}}{1 \mathrm{~mole} \mathrm{H}_{2}} \approx 2.14 \times 10^{7} \mathrm{~g}\] To convert to metric tons (kg), we divide by 1000: \[2.14 \times 10^{7} \mathrm{~g} \mathrm{H}_{2} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \approx 2.14 \times 10^{4} \mathrm{~kg}\]

Step 4: Determine moles of NaOH produced daily

From the balanced chemical equation, we know that for every one mole of Cl₂ produced, one mole of NaOH is also produced. So, the moles of NaOH produced daily is equal to the moles of Cl₂ produced daily (2.12 x 10⁷ moles).

Step 5: Calculate the mass of NaOH produced daily

We can now convert the moles of NaOH to mass by using the molar mass of NaOH (40.00 g/mol). \[2.12 \times 10^{7} \mathrm{~moles} \mathrm{NaOH} \times \frac{40.00 \mathrm{~g/mol}}{1 \mathrm{~mole} \mathrm{NaOH}} \approx 8.48 \times 10^{8} \mathrm{~g}\] To convert to metric tons (kg), we divide by 1000: \[8.48 \times 10^{8} \mathrm{~g} \mathrm{NaOH} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} \approx 8.48 \times 10^{5} \mathrm{~kg}\]

Step 6: Final quantities

In conclusion, using the stoichiometric calculations, we found that the chemical plant produces daily: - About 2.14 x 10⁴ kg (21.4 metric tons) of H₂. - About 8.48 x 10⁵ kg (848 metric tons) of NaOH.

Key Concepts

Stoichiometry in Electrolysis

Stoichiometry is an essential concept in chemistry that helps us calculate the quantities of reactants and products involved in chemical reactions. It involves using balanced chemical equations to relate the amounts of different substances. In our electrolysis problem, stoichiometry gives us proportionate relationships, like how two moles of sodium chloride (\(\mathrm{NaCl}\)) and two moles of water (\(\mathrm{H}_{2}\mathrm{O}\)) yield two moles of sodium hydroxide (\(\mathrm{NaOH}\)), one mole of hydrogen (\(\mathrm{H}_{2}\)), and one mole of chlorine (\(\mathrm{Cl}_{2}\)). From the balanced equation\(2\,\mathrm{NaCl}(aq) + 2\,\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\,\mathrm{NaOH}(aq) + \mathrm{H}_{2}(g) + \mathrm{Cl}_{2}(g)\), it becomes evident that precise ratios govern the reaction.

• 2 moles of \(\mathrm{NaCl}\) react with 2 moles of \(\mathrm{H}_{2}\mathrm{O}\).
• This results in the formation of \(\mathrm{NaOH}\) in a 1:1 ratio with \(\mathrm{Cl}_{2}\)
• \(\mathrm{H}_{2}\) generation is half of \(\mathrm{Cl}_{2}\) production.

Understanding these ratios allows us to predict the amounts of products formed when we know the quantity of one reactant.

Understanding Molar Mass

Molar mass is the mass of one mole of a substance and is expressed in grams per mole (\(\mathrm{g/mol}\)). Molar masses of elements and compounds are vital for conversion between mass and moles. For example, in the electrolysis reaction problem, we calculated the number of moles from a given mass using molar masses:\(\mathrm{Cl}_{2}\) has a molar mass of 70.90 \(\mathrm{g/mol}\), and \(\mathrm{H}_{2}\) has a molar mass of 2.02 \(\mathrm{g/mol}\). These were used to convert:

• Mass of \(\mathrm{Cl}_{2}\)
• Convert mass of \(\mathrm{H}_{2}\)

Knowing these values allows us to efficiently convert between measures, hence crucial for planning and conducting chemical reactions.

The Role of Chemical Equations

Chemical equations are symbolic representations of chemical reactions. They show the reactants transforming into products, often showcasing the relative amounts. In a balanced chemical equation like ours, each side has the same number of atoms of each element. Equations like\(2\,\mathrm{NaCl}(aq) + 2\,\mathrm{H}_{2}\mathrm{O}(l) \rightarrow 2\,\mathrm{NaOH}(aq) + \mathrm{H}_{2}(g) + \mathrm{Cl}_{2}(g)\), help ensure atoms' conservation, reflecting reality in reactions.
Balanced equations determine:

• Quantities of reactants & products.
• Stoichiometric proportions.
• Guidelines for laboratory and industrial processes.

Thus, they are the foundation of reaction stoichiometry and calculations we perform.

Conversion Calculations

Conversion calculations in chemistry allow us to move between different unit types, crucial for expressing reactant and product quantities. In our problem, we converted \(\mathrm{Cl}_{2}\) from kilograms to grams before calculating moles, essential for further stoichiometric calculations. This fundamental skill in chemistry involves steps like:

• Converting mass units: \(1.5\times 10^{6}\,\mathrm{kg}\) to \(1.5 \times 10^{9}\,\mathrm{g}\).
• Using molar mass to find moles.
• Converting back to different unit measures, e.g., grams to metric tons.

These calculations ensure accuracy and compatibility of chemical data across diverse applications, from academic exercises to industrial processes.