Problem 76 - General Chemistry: Principles and Modern Applications

Question

Nitrogen gas, \(\mathrm{N}_{2}\), can be prepared by passing gaseous ammonia over solid copper(II) oxide, \(\mathrm{CuO}\), at high temperatures. The other products of the reaction are solid copper, \(\mathrm{Cu},\) and water vapor. In a certain experiment, a reaction mixture containing \(18.1 \mathrm{g} \mathrm{NH}_{3}\) and \(90.4 \mathrm{g}\) CuO yields \(6.63 \mathrm{g} \mathrm{N}_{2}\). Calculate the percent yield for this experiment.

Step-by-Step Solution

Answer

The percent yield for this experiment is 44.5%

Step 1: Write down the balanced chemical equation of the reaction

First, formulate the balanced chemical equation of the reaction. The reaction given is: \[2NH_3(g) + 3CuO(s) \rightarrow N_2(g) + 3Cu(s) + 3H_2O(g)\]

Step 2: Identify the limiting reactant

Calculate the mole of each reactant. The molar mass of \(NH_3\) is approximately 17 g/mol, and that of \(CuO\) is approximately 80 g/mol. This means there are \( 18.1g \div 17 g/mol = 1.065 mol \) of \(NH_3\) and \( 90.4g \div 80 g/mol = 1.13 mol \) of \(CuO\). Given the stoichiometry of the reaction (2:3 ratio), \(NH_3\) is the limiting reactant because it will be consumed completely first.

Step 3: Calculate the theoretical yield

With NH3 as the limiting reactant, calculate the amount of N2 that could be theoretically produced. Based on the stoichiometric ratio from the chemical reaction, 2 mol of NH3 will produce 1 mol of N2. Thus, the theoretical yield of N2 is \(1.065 mol \div 2 = 0.5325 mol\). The molar mass of N2 is approximately 28 g/mol, so in grams, the theoretical yield is \( 0.5325 mol \times 28g/mol = 14.91 g\).

Step 4: Calculate the percent yield

To compute the percent yield, the formula is: \(Percent yield = \frac{Actual yield}{Theoretical yield} \times 100\%\). Plugging in the values, the actual yield of nitrogen gas, N2 (from problem statement) is 6.63g and the theoretical yield we calculated to be 14.91g. Thus, the percent yield of the reaction is \( \frac{6.63g}{14.91g} \times 100\% = 44.5\% \)

Key Concepts

Percent Yield

Percent yield is a measure of the efficiency of a chemical reaction. It tells us how much product was actually obtained compared to the maximum possible amount. Understanding percent yield is essential.
The actual yield is the amount of product obtained from the experiment, which can be different from the theoretical yield.
Theoretical yield is the calculated amount expected based on stoichiometry.

• The formula to calculate percent yield is: \[ \text{Percent yield} = \left(\frac{\text{Actual yield}}{\text{Theoretical yield}}\right) \times 100\%. \]

In this exercise, the theoretical yield of nitrogen gas was calculated to be 14.91 grams.
The actual yield obtained was 6.63 grams.
Hence, the percent yield calculated was 44.5\%. A lower percent yield suggests inefficiencies or loss of materials during the reaction process or other experimental errors. Understanding and improving percent yield can help make chemical reactions more efficient and cost-effective.

Limiting Reactant

The concept of limiting reactant is crucial in determining how much product a reaction will yield. In any chemical reaction, the limiting reactant is the substance that is completely consumed first.
This dictates the maximum amount of product that can form because no more product can be made once the limiting reactant is used up.

• To identify the limiting reactant, compare the mole ratio of the reactants provided to the mole ratio in the balanced equation.
• In the given reaction, ammonia ( \( \text{NH}_3 \) ) and copper(II) oxide ( \( \text{CuO} \) ) were given.
• The balances equation ratio is 2:3 (\(\text{NH}_3\) to \(\text{CuO}\)).
• Upon calculation, ammonia was used as the limiting reactant.
• The moles of \(\text{NH}_3\) were calculated as 1.065, and \(\text{CuO}\) was 1.13.
• Since more than 3 moles of \(\text{NH}_3\) are needed to consume all \(\text{CuO}\), \(\text{NH}_3\) limits the reaction progress.

Recognizing the limiting reactant helps in calculating the theoretical yield and designing processes for optimal resource usage.

Balanced Chemical Equation

A balanced chemical equation is foundational for grasping how chemical reactions occur and are represented. This balance ensures that both mass and charge are conserved in a reaction.
Every chemical reaction must be balanced so that the same number of atoms of each element are present in both the reactants and the products.

• In the given reaction: \[2NH_3(g) + 3CuO(s) \rightarrow N_2(g) + 3Cu(s) + 3H_2O(g)\],
• The coefficients indicate the number of moles of each substance involved.
• This balancing helps ascertain stoichiometric relationships, enabling calculations such as determining the limiting reactant and theoretical yield.

Understanding the balanced equation enables students to predict the quantities of reactants required.
It also allows prediction of the amount of product formed in a chemical reaction.