Problem 20 - Chemistry and Chemical Reactivity

Question

Ammonia gas can be prepared by the following reaction: $$\begin{aligned} \mathrm{CaO}(\mathrm{s})+2 \mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s}) & \rightarrow \\ 2 \mathrm{NH}_{3}(\mathrm{g}) &+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})+\mathrm{CaCl}_{2}(\mathrm{s}) \end{aligned}$$ 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 NH\(_3\) is 23.97%.

Step 1: Understand the Percent Yield Formula

Percent yield is calculated using the formula: \( \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \)%.

Step 2: Identify Actual and Theoretical Yield

The actual yield of NH\(_3\) given in the problem is 16.3 g. The theoretical yield of NH\(_3\) is provided as 68.0 g.

Step 3: Substitute Values into the Formula

Use the values for actual and theoretical yield in the percent yield formula: \( \text{Percent Yield} = \left( \frac{16.3 \text{ g}}{68.0 \text{ g}} \right) \times 100 \)%.

Step 4: Calculate the Percent Yield

Perform the division: \( \frac{16.3}{68.0} = 0.2397 \). Then multiply by 100 to convert to a percentage: \( 0.2397 \times 100 = 23.97 \% \).

Key Concepts

Theoretical Yield

Theoretical yield refers to the maximum amount of product that can be generated from a given set of reactants, assuming a perfectly efficient chemical reaction where everything reacts completely. It is deduced based on stoichiometry, which is a subfield of chemistry that involves calculating the quantitative relationships in chemical reactions.
This calculation takes into account the molecular weights of reactants and products as well as the balanced chemical equation, which indicates the proportions in which different substances react.
A balanced equation is crucial for finding the theoretical yield. In this exercise, the balanced equation allows us to determine that from certain masses of \( \text{CaO} \) and \( \text{NH}_{4}\text{Cl} \), we should ideally produce \( 68.0 \text{ g} \) of \( \text{NH}_3 \).
However, keep in mind that theoretical yield is an ideal value and is rarely achieved in real-world scenarios due to several types of losses.

Actual Yield

The actual yield is the amount of product that is actually formed when the reaction is carried out in practice. It is observed and recorded from the experiment you conduct.
In chemical reactions, the actual yield can often be less than the theoretical yield due to various factors such as incomplete reactions, side reactions, or loss of material during the process.
In the context of the exercise, the actual yield of \( \text{NH}_{3} \) was \( 16.3\text{ g} \).
It is important to accurately measure the actual yield to calculate the percent yield, which helps us gauge the efficiency of the reaction. Observing large discrepancies between actual and theoretical yields can indicate the need for process improvements.

Chemical Reactions

Chemical reactions involve the transformation of substances through the breaking and forming of bonds to convert reactants into products. These processes are guided by the laws of conservation, specifically mass and energy conservation.
In the provided exercise, the reaction between \( \text{CaO} \) and \( \text{NH}_{4}\text{Cl} \) results in products like \( \text{NH}_{3} \) and \( \text{CaCl}_{2} \).
The underlying principle of balancing chemical equations ensures that the number of atoms for each element remains constant throughout the reaction.

• Chemical reactions can be influenced by several factors including temperature, pressure, and concentration of reactants.
• Understanding the nature of the reactants and the conditions of the reaction can help optimize yields and reduce waste.

Improving reaction conditions often leads to higher yields and can make laboratory or industrial processes more sustainable.