Problem 63 - Chemistry and Chemical Reactivity

Question

When an electric current is passed through an aqueous solution of \(\mathrm{NaCl}\), the valuable industrial chemicals \(\mathrm{H}_{2}(\mathrm{g}), \mathrm{Cl}_{2}(\mathrm{g}),\) and \(\mathrm{NaOH}\) are produced. $$2 \mathrm{NaCl}(\mathrm{aq})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow \mathrm{H}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})+2 \mathrm{NaOH}(\mathrm{aq})$$ What mass of NaOH can be formed from 15.0 L of 0.35 M NaCl? What mass of chlorine is obtained?

Step-by-Step Solution

Answer

210 g of NaOH and 186.61 g of Cl2 are produced.

Step 1: Determine Moles of NaCl

First, we need to calculate the moles of NaCl in the solution. We use the formula for moles: \( ext{Moles} = ext{Molarity} \times ext{Volume} \). Since we have 0.35 M NaCl and a volume of 15.0 L, we find the moles as follows: \( ext{Moles of NaCl} = 0.35 \, M \times 15.0 \, L = 5.25 \, ext{moles} \).

Step 2: Use Stoichiometry for NaOH

The balanced chemical equation shows that 2 moles of NaCl produce 2 moles of NaOH. Therefore, the moles of NaOH will be the same as the moles of NaCl, which is 5.25 moles.

Step 3: Calculate Mass of NaOH

The formula to calculate mass is \( ext{Mass} = ext{Moles} \times ext{Molar Mass} \). The molar mass of NaOH is approximately 40.00 g/mol. Therefore, the mass of NaOH formed is \( 5.25 \, ext{moles} \times 40.00 \, ext{g/mol} = 210 \, ext{g} \).

Step 4: Use Stoichiometry for Chlorine Gas

According to the balanced equation, 2 moles of NaCl produce 1 mole of \( \mathrm{Cl}_{2} \). This means from 5.25 moles of NaCl, we can produce \( \frac{5.25}{2} = 2.625 \, ext{moles of} \, \mathrm{Cl}_{2} \).

Step 5: Calculate Mass of Chlorine Gas

The molar mass of \( \mathrm{Cl}_{2} \) is approximately 70.90 g/mol. Therefore, the mass of chlorine gas produced is \( 2.625 \, ext{moles} \times 70.90 \, ext{g/mol} = 186.61 \, ext{g} \).

Key Concepts

Understanding Molarity

Molarity is a way to express the concentration of a solution. It's calculated as the number of moles of solute per liter of solution. For example, in the given problem, the solution has a molarity of 0.35 M, meaning there are 0.35 moles of sodium chloride (NaCl) per liter of solution.

To find out how many moles you have when given the volume of the solution, simply multiply the molarity by the solution's volume in liters. In this case, with 15 liters of 0.35 M NaCl, the calculation is:

• Moles of NaCl = 0.35 M × 15 L = 5.25 moles

Understanding molarity helps you determine how much of a chemical is present in a given volume, which is crucial in reactions like electrolysis.

Basics of Stoichiometry

Stoichiometry is the study of the quantitative relationships in chemical reactions. It helps us figure out how much of each substance is consumed and produced during a reaction.

In the exercise, the balanced equation is:

• 2 NaCl + 2 H\(_2\)O → H\(_2\) + Cl\(_2\) + 2 NaOH

This tells us:

• 2 moles of NaCl will produce 2 moles of NaOH.
• 2 moles of NaCl will produce 1 mole of Cl\(_2\).

Knowing these relationships allows us to calculate the amount of products formed. For instance, from 5.25 moles of NaCl:

• You can make 5.25 moles of NaOH because of the 1:1 ratio.
• You can produce 2.625 moles of Cl\(_2\) because of the 2:1 ratio.

Stoichiometry is like a recipe for chemistry, ensuring you know the right proportions of ingredients.

Interpreting Chemical Equations

Chemical equations provide a symbolic representation of chemical reactions. They're balanced to conserve mass and show the correct proportions of reactants and products.

The equation given in the problem:

• 2 NaCl(aq) + 2 H\(_2\)O(l) → H\(_2\)(g) + Cl\(_2\)(g) + 2 NaOH(aq)

This tells us that:

• 2 sodium chloride molecules are needed for the reaction to proceed.
• Water is involved but isn't in the products as a separate entity.
• The reactants result in the formation of hydrogen gas, chlorine gas, and sodium hydroxide.

Balancing equations ensures no atoms are lost or gained, maintaining the law of conservation of mass. This balance helps understand the moles and mass of each substance involved.