Problem 84 - Chemistry: An Atoms-Focused Approach

Question

Baking soda (NaHCO \(_{3}\) ) is produced on an industrial scale by the Solvay process. A key reaction in the process is $$\mathrm{NaCl}(a q)+\mathrm{NH}_{3}(a q)+\mathrm{CO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow$$ $$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mathrm{NaHCO}_{3}(s)+\mathrm{NH}_{4} \mathrm{Cl}(a q)$$ Suppose a reaction vessel initially contains \(58.5 \mathrm{kg} \mathrm{NaCl}\) \(18.8 \mathrm{kg} \mathrm{NH}_{3},\) and excess \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) If \(66 \mathrm{kg} \mathrm{NaHCO}_{3}\) is produced, what is the percent yield?

Step-by-Step Solution

Answer

Answer: The percent yield of NaHCO3 in this reaction is approximately 78.56%.

Step 1: Write down the balanced chemical equation

The balanced chemical equation is already provided: NaCl(aq) + NH3(aq) + CO2(aq) + H2O(l) -> NaHCO3(s) + NH4Cl(aq)

Step 2: Calculate the moles of the reactants

We are given the mass of NaCl (58.5 kg) and NH3 (18.8 kg). We can convert the mass of each reactant to moles using their respective molar mass: Molar mass of NaCl = 58.44 g/mol Molar mass of NH3 = 17.03 g/mol Moles of NaCl = (58.5 kg * 1000 g/kg) / 58.44 g/mol = 1000 moles; (1 kg = 1000 g) Moles of NH3 = (18.8 kg * 1000 g/kg) / 17.03 g/mol = 1104.52 moles

Step 3: Find the limiting reactant

We can find the limiting reactant by checking the mole ratio between the reactants and the product, according to the balanced equation. In this case, the ratio is 1:1 for NaCl and NH3 to NaHCO3: Moles of NaCl / Moles of NH3 = 1000 moles / 1104.52 moles = 0.905 (approximately) Since the ratio is less than 1, NaCl is the limiting reactant.

Step 4: Calculate the theoretical yield of NaHCO3

From the balanced chemical equation, we can see that 1 mole of NaCl produces 1 mole of NaHCO3. Therefore, the theoretical yield of NaHCO3 based on the moles of the limiting reactant (NaCl) is: Moles of NaHCO3 = 1000 moles (same as moles of NaCl) Now we can convert the moles of NaHCO3 to mass, using its molar mass: Molar mass of NaHCO3 = 84.01 g/mol Theoretical yield of NaHCO3 = 1000 moles * 84.01 g/mol = 84010 g or 84.01 kg

Step 5: Calculate the actual yield of NaHCO3

The actual yield of NaHCO3 is given in the problem: 66 kg.

Step 6: Calculate the percent yield

Percent yield is calculated using the following formula: Percent yield = (Actual yield / Theoretical yield) * 100 Percent yield = (66 kg / 84.01 kg) * 100 = 78.56% The percent yield of NaHCO3 in this reaction is approximately 78.56%.

Key Concepts

Balancing Chemical Equations

In a chemical reaction, balancing chemical equations is crucial to ensure that the law of conservation of mass is honored. This law states that matter cannot be created or destroyed in an isolated system. Hence, the number of atoms for each element must remain constant, before and after the reaction. For example, in the Solvay process, the equation:

• \[\mathrm{NaCl} + \mathrm{NH_3} + \mathrm{CO_2} + \mathrm{H_2O} \rightarrow \mathrm{NaHCO_3} + \mathrm{NH_4Cl}\]

is already balanced. This means that the same number of atoms of each element appear on both sides of the equation. Balancing ensures that the stoichiometric coefficients (the numbers before molecules in a balanced equation) accurately reflect the proportional amounts of substances involved, which is essential for further calculations such as identifying limiting reactants and determining theoretical yield.

Limiting Reactant

The concept of a limiting reactant is pivotal when dealing with chemical reactions, especially in industrial processes. The limiting reactant is the substance that is entirely consumed first in a chemical reaction. This reactant determines the maximum amount of product that can form.
In the provided example, we compare moles of sodium chloride (NaCl) and ammonia (NH₃). With 1000 moles of NaCl and 1104.52 moles of NH₃, NaCl is the limiting reactant because you would run out of it first if the reaction were to proceed to completion.
This is calculated using the given mass and determining the number of moles by dividing by the molar mass. Understanding which reactant limits the reaction helps in calculating the theoretical yield, as it sets a cap on how much product you can produce.

Percent Yield

Percent yield is a valuable metric in chemical manufacturing that measures the efficiency of a reaction. It compares the actual yield obtained from the reaction to the theoretical yield calculated based on the stoichiometry of the balanced equation.
The formula used is:

• \[\text{Percent yield} = \left( \frac{\text{Actual yield}}{\text{Theoretical yield}} \right) \times 100\]

In our context of the Solvay process, the actual product mass of sodium bicarbonate (NaHCO₃) was 66 kg, while the theoretical yield was 84.01 kg. Using these figures, the percent yield of the reaction was calculated to be approximately 78.56%. A high yield is desirable, showing that most of the reactants converted into the desired product, but in real-world scenarios, anything over 70% is often considered quite efficient.

Molar Mass Calculation

Calculating molar mass is fundamental when converting between mass and moles, an essential step in problem-solving for stoichiometry in chemistry. The molar mass of a compound can be found by summing the atomic masses of its constituent elements, which are typically given on the periodic table in units of grams per mole.
For instance, the molar mass of sodium chloride (NaCl) is 58.44 g/mol, derived from adding the molar masses of sodium (Na) and chlorine (Cl). Similarly, for ammonia (NH₃), the molar mass is 17.03 g/mol. These values are crucial for converting the mass of substances into a quantity of moles, which are then used to perform further calculations like determining the limiting reactant and theoretical yields.

• Understanding molar mass helps in accurately measuring and converting amounts when mixing reactants to ensure the right proportions for optimal results in any chemical reaction.