Problem 94 - Chemistry: The Central Science

Question

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

Given the balanced chemical equation: \(2\mathrm{NaCl}(aq) + 2\mathrm{H_2O}(l) \rightarrow 2\mathrm{NaOH}(aq) + \mathrm{H_2}(g) + \mathrm{Cl_2}(g)\), and the production of \(1.5 \times 10^{6}\,\mathrm{kg}\) of \(\mathrm{Cl}_{2}\) daily, we can estimate the daily production of \(\mathrm{H}_2\) and \(\mathrm{NaOH}\) as follows: 1. Determine the mole ratio: \(2\,\mathrm{NaOH} : 1\,\mathrm{H_2} : 1\,\mathrm{Cl_2}\) 2. Convert mass of \(\mathrm{Cl_2}\) to moles: \(n_{\mathrm{Cl_{2}}} = \frac{1.5 \times 10^{6} \times 1000\,\mathrm{g}}{70.90\,\mathrm{g/mol}}\) 3. Estimate moles of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\): \(n_{\mathrm{H_{2}}} = n_{\mathrm{Cl_{2}}}\) and \(n_{\mathrm{NaOH}} = 2 \times n_{\mathrm{Cl_{2}}}\) 4. Convert moles of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\) to mass: - Mass of \(\mathrm{H_2}\): \(m_{\mathrm{H_2}} = n_{\mathrm{H_2}} \times 2.016\,\mathrm{g/mol} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}}\) - Mass of \(\mathrm{NaOH}\): \(m_{\mathrm{NaOH}} = n_{\mathrm{NaOH}} \times 39.997\,\mathrm{g/mol} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}}\) Plugging the values from Step 2 into Step 4, we find that the plant produces approximately \(4.29 \times 10^5\,\mathrm{kg}\) of \(\mathrm{H_2}\) and \(8.58 \times 10^5\,\mathrm{kg}\) of \(\mathrm{NaOH}\) daily.

Step 1: Determine the mole ratio of \(\mathrm{H_{2}}\), \(\mathrm{Cl_{2}}\), and \(\mathrm{NaOH}\)

From the balanced chemical equation, we can see the following mole ratio: \(2\,\mathrm{NaOH} : 1\,\mathrm{H_2} : 1\,\mathrm{Cl_2}\)

Step 2: Convert the mass of \(\mathrm{Cl_{2}}\) produced daily into moles

We are given that the plant produces \(1.5 \times 10^{6}\,\mathrm{kg}\) of \(\mathrm{Cl_{2}}\) daily. To convert this mass into moles, we divide it by the molar mass of \(\mathrm{Cl_{2}}\): Molar mass of \(\mathrm{Cl_{2}} = 2 \times 35.45\,\mathrm{g/mol} = 70.90\,\mathrm{g/mol}\) Since \(1\,\mathrm{kg} = 1000\,\mathrm{g}\), the mass of \(\mathrm{Cl_{2}}\) produced daily in grams is \(1.5 \times 10^{6} \times 1000\,\mathrm{g}\). Therefore, the number of moles of \(\mathrm{Cl_{2}}\) produced daily is: \[ n_{\mathrm{Cl_{2}}} = \frac{1.5 \times 10^{6} \times 1000\,\mathrm{g}}{70.90\,\mathrm{g/mol}} \]

Step 3: Estimate the moles of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\) produced daily

Using the mole ratio (Step 1), we can now estimate the moles of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\) produced daily. For every mole of \(\mathrm{Cl_{2}}\) produced, 1 mole of \(\mathrm{H_{2}}\) and 2 moles of \(\mathrm{NaOH}\) are produced. Therefore, \[n_{\mathrm{H_{2}}} = n_{\mathrm{Cl_{2}}}\] and \[n_{\mathrm{NaOH}} = 2 \times n_{\mathrm{Cl_{2}}}\]

Step 4: Convert the moles of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\) back into mass (kg)

To find the mass of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\) produced daily, we multiply their respective moles by their molar masses and convert back to kg. Molar mass of \(\mathrm{H_{2}} = 2.016\,\mathrm{g/mol}\) Mass of \(\mathrm{H_2}\) produced daily: \[m_{\mathrm{H_2}} = n_{\mathrm{H_2}} \times 2.016\,\mathrm{g/mol} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}}\] Molar mass of \(\mathrm{NaOH} = 22.99 + 15.999 + 1.008\,\mathrm{g/mol} = 39.997\,\mathrm{g/mol}\) Mass of \(\mathrm{NaOH}\) produced daily: \[m_{\mathrm{NaOH}} = n_{\mathrm{NaOH}} \times 39.997\,\mathrm{g/mol} \times \frac{1\,\mathrm{kg}}{1000\,\mathrm{g}}\] Now we can plug the values from Step 2 into Step 4 and estimate the quantities of \(\mathrm{H_2}\) and \(\mathrm{NaOH}\) produced daily.

Key Concepts

Mole Ratio

The concept of mole ratio is crucial in understanding chemical reactions, as it links the amounts of reactants to the products. In a balanced chemical equation, the coefficients represent the mole ratios of the substances involved. For the electrolysis of \(\mathrm{NaCl}\), the equation is: \[2\,\mathrm{NaCl}(aq) + 2\,\mathrm{H_2O}(l) \rightarrow 2\,\mathrm{NaOH}(aq) + \mathrm{H_2}(g) + \mathrm{Cl_2}(g)\]From this, you can see that:- 2 moles of \(\mathrm{NaOH}\) are produced for every 1 mole each of \(\mathrm{H_2}\) and \(\mathrm{Cl_2}\).- The ratio is \(2 : 1 : 1\) for \(\mathrm{NaOH} : \mathrm{H_2} : \mathrm{Cl_2}\).Understanding this ratio allows you to proportionally calculate the amounts of other substances formed when given the amount of one substance in a reaction.

Chemical Equations

Chemical equations are fundamental to understanding reactions. They describe transformations from reactants to products, detailing reactants on the left side and products on the right side. The equation must be balanced to show the conservation of mass, meaning the number of atoms for each element involved must be equal on both sides.For instance, in the decomposition of \(\mathrm{NaCl}\) through electrolysis:- Starts with \(2\,\mathrm{NaCl}\) and \(2\,\mathrm{H_2O}\)- Produces \(2\,\mathrm{NaOH}\), \(\mathrm{H_2}\), and \(\mathrm{Cl_2}\)Each part of the equation corresponds to a particular change, reflecting real-world chemical changes at a microscopic level. This balancing ensures that you can accurately use equations to calculate quantities produced in reactions.

Mass to Moles Conversion

Converting mass to moles is an essential step in stoichiometry, as chemical equations provide ratios in terms of moles. To perform this conversion, you use the formula:\[ n = \frac{m}{M} \]where:- \(n\) is the number of moles,- \(m\) is the mass in grams,- \(M\) is the molar mass in \(\mathrm{g/mol}\).For example, when converting the mass of \(\mathrm{Cl_2}\) produced, we first change kilograms to grams (since \(1\,\mathrm{kg} = 1000\,\mathrm{g}\)), and then divide by its molar mass, \(70.90\,\mathrm{g/mol}\), to find the number of moles. This step is crucial in transitioning from a tangible mass to an abstract concept of moles for further calculations.

Molar Mass Calculation

Calculating the molar mass of a compound lets you link the compound's mass to its moles, forming the basis of converting between mass and moles. To find the molar mass:

• Identify each element in the chemical formula.
• Use the periodic table to find the atomic masses of these elements.
• Sum the atomic masses, accounting for the number of each type of atom in the compound.

Taking \(\mathrm{Cl_2}\) as an example, you have two chlorine atoms, each with a mass of \(35.45\,\mathrm{g/mol}\). The molar mass is therefore:\[\mathrm{Cl_2}: 2 \times 35.45 = 70.90\,\mathrm{g/mol}\]Understanding how to perform these calculations extends to any compound, including \(\mathrm{NaOH}\), where you'd sum the masses of Na, O, and H for an accurate molar mass used in further stoichiometric calculations.