Problem 159 - Introductory Chemistry Atoms First

Question

Consider the unbalanced chemical equation \(\mathrm{CaCN}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3}+\mathrm{NH}_{3}\) If you began the reaction with \(5.65 \mathrm{~g}\) of \(\mathrm{CaCN}_{2}\) and \(12.2 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\), how many grams of \(\mathrm{NH}_{3}\) would be produced if the reaction had an \(86.0 \%\) yield?

Step-by-Step Solution

Answer

The balanced chemical equation is given by: \( CaCN_2 + 2H_2O \rightarrow CaCO_3 + 2NH_3 \). In this reaction, the limiting reactant is CaCN₂, and the theoretical yield of NH₃ is 2.09 g. Given an 86.0% yield, the actual amount of NH₃ produced would be approximately 1.80 g.

Step 1: Balance the chemical equation

To balance the equation, we need to make sure that the number of atoms of each element on the reactants side is equal to the number of atoms of each element on the products side. The balanced chemical equation: \( CaCN_2 + 2H_2O \rightarrow CaCO_3 + 2NH_3 \) Now, the equation is balanced, and we can proceed to finding the limiting reactant.

Step 2: Find the limiting reactant

To find the limiting reactant, we will first find the molar mass of each reactant and convert grams to moles for each. Next, we will use the mole ratio (stoichiometry) from the balanced chemical equation to determine the limiting reactant. Molar masses: - CaCN2: Ca (40.08 g/mol) + 2C (24.02 g/mol) + 2N (28.02 g/mol) = 92.12 g/mol - H2O: 2H (2.02 g/mol) + O (16.00 g/mol) = 18.02 g/mol Moles of reactants: - Moles of CaCN2: 5.65 g / 92.12 g/mol ≈ 0.0613 mol - Moles of H2O: 12.2 g / 18.02 g/mol ≈ 0.677 mol Now, we will use the mole ratio in the balanced equation to determine the limiting reactant. As the ratio is 1 mole of CaCN2 : 2 moles of H2O, we will divide the number of moles of each reactant by the stoichiometric coefficients. - CaCN2: 0.0613 mol / 1 = 0.0613 - H2O: 0.677 mol / 2 ≈ 0.3385 Since 0.0613 is smaller than 0.3385, CaCN2 is the limiting reactant.

Step 3: Calculate the theoretical yield of NH3

Now that we know CaCN2 is the limiting reactant, we can calculate the theoretical yield of NH3 using the mole ratio. The ratio in the balanced equation is 1 mole of CaCN2 : 2 moles of NH3. To find how many moles of NH3 would be produced, we have 0.0613 mol of CaCN2 × (2 mol NH3 / 1 mol CaCN2) ≈ 0.1226 mol of NH3. Now, let's convert moles of NH3 to grams using the molar mass of NH3: Molar mass of NH3: N (14.01 g/mol) + 3H (3.03 g/mol) = 17.04 g/mol Grams of NH3: 0.1226 mol × 17.04 g/mol ≈ 2.09 g (theoretical yield)

Step 4: Calculate the actual yield of NH3 using the percentage yield

Now we will use the given 86.0% yield to calculate the actual amount of NH3 produced: Actual yield of NH3: 2.09 g × 86.0% = 2.09 g × 0.86 ≈ 1.80 g Therefore, 1.80 grams of NH3 would be produced in the reaction with an 86.0% yield.

Key Concepts

Chemical Equation Balancing

When dealing with chemical reactions, the very first step is to ensure the chemical equation is balanced. Balancing a chemical equation means making sure that the same number of each type of atom appears on both sides of the equation. For our example reaction, we have\( \mathrm{CaCN}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3}+\mathrm{NH}_{3} \).

Initially, the atoms on both sides are not equal, which makes our equation unbalanced. To balance it, you will need to adjust the coefficients (the numbers in front of each molecule) to ensure each type of atom is balanced.

• Start by counting the atoms of each element in the reactants and products.
• Adjust the coefficients step by step to make the number of each type of atom identical on both sides. For instance, if you have 2 hydrogens on one side but 4 on the other, adjust the coefficients to make them equal.
• Our balanced equation becomes: \( \mathrm{CaCN}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2 \mathrm{NH}_{3} \).

This balanced equation is crucial for calculating other key aspects of the reaction, such as the limiting reactant.

Molar Mass Calculation

Molar mass is a critical concept in chemistry that represents the mass of one mole of a given substance. For each chemical involved, it is calculated by adding up the atomic masses of all the atoms in its formula. Here's how you find these:

• Start with the chemical formula. For \(\mathrm{CaCN}_{2}\), the molar mass is calculated as follows: - Calcium (Ca) has an atomic mass of about 40.08 g/mol. - Carbon (C), which is \(2\) in the compound, has an atomic mass of about 12.01 g/mol, leading to \(2 \times 12.01 = 24.02\) g/mol. - Nitrogen (N) is also \(2\), adding up to \(2 \times 14.01 = 28.02\) g/mol. - Hence, the molar mass for \(\mathrm{CaCN}_{2}\) is \(40.08 + 24.02 + 28.02 = 92.12\) g/mol.
• For \(\mathrm{H}_{2} \mathrm{O}\): - Hydrogen (H) is \(2 \times 1.01 = 2.02\) g/mol. - Oxygen (O) is \(16.00\) g/mol. - The total molar mass becomes \(18.02\) g/mol.

Molar mass helps you to convert grams of the reactant into moles, which is the next step in solving stoichiometric problems.

Mole Ratio

The mole ratio is an essential part of stoichiometry. It helps us convert an initial amount of reactants into products using the balanced chemical equation. In our reaction:\( \mathrm{CaCN}_{2} + 2 \mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{CaCO}_{3} + 2 \mathrm{NH}_{3} \), the ratio tells us that:

• 1 mole of \(\mathrm{CaCN}_{2}\) produces 2 moles of \(\mathrm{NH}_{3}\).
• For \(\mathrm{H}_{2} \mathrm{O}\), every 2 moles can also produce 2 moles of \(\mathrm{NH}_{3}\).

Using the mole ratio helps us identify which reactant limits the reaction—that is, the reactant that runs out first and stops the reaction from proceeding. This is often termed the "limiting reactant."

To determine the limiting reactant:

• Calculate the moles of each reactant based on their initial masses.
• Compare them using the stoichiometric coefficients from the balanced equation (1 for \(\mathrm{CaCN}_{2}\) and 2 for \(\mathrm{H}_{2} \mathrm{O}\)).
• The smallest resulting value identifies your limiting reactant. In this case, \(\mathrm{CaCN}_{2}\) is the limiting reactant since it produces fewer moles of \(\mathrm{NH}_{3}\).

Percentage Yield Calculation

Chemical reactions in the real world do not always proceed perfectly. Percentage yield is a measure that tells us how effective a reaction actually is, compared to what was theoretically possible. It is calculated using the formula:\[ \text{Percentage Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \]

For this, calculate the actual and theoretical yields:

• Theoretical yield: Use the limiting reactant (in this case, \(\mathrm{CaCN}_{2}\)) and the mole ratio to find the maximum amount of product that could be formed—here, \(0.1226\) moles of \(\mathrm{NH}_{3}\) translate into \(2.09\) grams.
• Actual yield: Assume it's given or measured—in this problem, it's derived from an \(86\%\) yield, implying about \(1.80\) grams of \(\mathrm{NH}_{3}\) was actually produced.

By plugging these values into the percentage yield formula, the reaction's efficiency can be evaluated, which in our case, confirms \(86\%\) efficiency, revealing how close performance approaches theoretical expectations.