Problem 20 - Chemistry and Chemical Reactivity

Question

Ammonia gas can be prepared by the following reaction: \(\mathrm{CaO}(\mathrm{s})+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) \rightarrow\) $$ 2 \mathrm{NH}_{3}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CaCl}_{2}(\mathrm{s}) $$ If \(112 \mathrm{g}\) of \(\mathrm{CaO}\) and \(224 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) are mixed, the theoretical yield of \(\mathrm{NH}_{3}\) is \(68.0 \mathrm{g}\) (Study Question 12 ). If only \(16.3 \mathrm{g}\) of \(\mathrm{NH}_{3}\) is actually obtained, what is its percent yield?

Step-by-Step Solution

Answer

The percent yield of \(\text{NH}_3\) is approximately 24\%.

Step 1: Identify Known Quantities

We are given the theoretical yield of ammonia as \(68.0 \, \text{g}\) and the actual yield as \(16.3 \, \text{g}\). To find the percent yield, we'll use these values.

Step 2: Use the Percent Yield Formula

The formula for percent yield is \(\text{Percent Yield} = \left(\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\right) \times 100\%\). Insert the known values into the formula.

Step 3: Calculate the Percent Yield

Substitute the actual yield of \(16.3 \, \text{g}\) and the theoretical yield of \(68.0 \, \text{g}\) into the formula: \[\text{Percent Yield} = \left(\frac{16.3}{68.0}\right) \times 100\%\]Perform the division to find the proportion.

Step 4: Compute the Result

Calculate \(\frac{16.3}{68.0} = 0.2397\). Multiply by 100 to convert to a percentage: \[0.2397 \times 100 = 23.97\%\].

Key Concepts

Theoretical Yield

Theoretical yield represents the maximum amount of product that can be generated from a chemical reaction, based on the complete conversion of the limiting reactant. This is a calculated expectation, determined using the balanced chemical equation and stoichiometry.
Calculating the theoretical yield involves several steps:

• Identify the balanced chemical equation. Here, we use \( \mathrm{CaO} + 2\ \mathrm{NH}_{4} \mathrm{Cl} \rightarrow 2\ \mathrm{NH}_{3} + \mathrm{H}_{2} \mathrm{O} + \mathrm{CaCl}_{2} \).
• Determine the limiting reactant. This is the reactant that will run out first, limiting the amount of product formed.
• Use stoichiometry to calculate the amount of product formed from the limiting reactant. Apply molar ratios from the balanced equation to determine the theoretical yield.

Understanding the theoretical yield is key to evaluating how efficient a reaction is likely to be under ideal conditions.

Actual Yield

Actual yield refers to the real amount of product obtained from a reaction. This is often less than the theoretical yield due to various factors such as incomplete reactions, impurities, or measurement errors.
Achieving the actual yield involves:

• Conducting the chemical reaction as accurately as possible.
• Measuring the product obtained carefully to ensure precision.

Several reasons might lead the actual yield to differ from the theoretical yield:

• Side reactions that consume some of the reactants.
• Loss of product during collection or transfer.
• Unexpected environmental factors affecting reaction dynamics.

Understanding the actual yield helps chemists and students acknowledge real-world limitations in experiments.

Chemical Reaction Stoichiometry

Chemical reaction stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It provides the coefficients that dictate the proportions of reactants needed to form products.
To utilize stoichiometry:

• Examine the balanced chemical equation.
• Use the molar ratios to calculate how much of each reactant is required or how much product will be generated.
• Apply these ratios to convert between masses, moles, and volumes (if dealing with gases).

For example, in our reaction, stoichiometry tells us that 2 moles of \( \mathrm{NH}_{4} \mathrm{Cl} \) produce 2 moles of \( \mathrm{NH}_{3} \). By understanding these ratios, it’s easier to solve problems related to the yield and efficiency of reactions. Accurate stoichiometric calculations are essential for predicting yield and scaling reactions in industrial applications.