Problem 84 - Introductory Chemistry Atoms First

Question

Chlorine \(\left(\mathrm{Cl}_{2}\right)\) and fluorine \(\left(\mathrm{F}_{2}\right)\) react to form \(\mathrm{ClF}_{3}\). A reaction vessel contains \(10.00 \mathrm{~g} \mathrm{Cl}_{2}\) and \(10.00 \mathrm{~g} \mathrm{~F}_{2}\). (Hint: Refer to Problem 9.82.) (a) Write a balanced chemical equation for this reaction. (b) Which reactant is limiting? (c) What is the theoretical yield for this reaction in grams? (d) How many grams of excess reactant are left over at the end of the reaction? (e) When this reaction is actually performed, \(12.50 \mathrm{~g}\) of \(\mathrm{ClF}_{3}\) is recovered. What is the percent yield of the reaction?

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction between chlorine and fluorine to form ClF3 is: Cl2 + 3F2 → 2ClF3. The limiting reactant is F2, the theoretical yield of ClF3 is 48.61 g, there are 3.79 g of excess Cl2 left over at the end of the reaction, and the percent yield of the reaction is approximately 25.72%.

Step 1: Write a balanced chemical equation for the reaction

Firstly, let's write down the unbalanced chemical equation for the reaction of chlorine and fluorine to form ClF3: Cl2 + F2 → ClF3 Now, balance the equation by placing the balanced coefficients in front of the reactants and products: \( \) \( \) 1/2 Cl2 + 3/2 F2 → ClF3 The balanced chemical equation is: \( \) \( \) Cl2 + 3F2 → 2ClF3

Step 2: Determine the limiting reactant

Next, we need to find the limiting reactant. We'll use the initial amounts of each reactant and their molar masses to determine the number of moles of each reactant. Calculate the moles of reactants given their masses (10 g each) and molar masses (35.45 g/mol for Cl2 and 38.00 g/mol for F2): moles of Cl2 = (10.00 g) / (35.45 g/mol) = 0.282 mol moles of F2 = (10.00 g) / (38.00 g/mol) = 0.263 mol Now divide the moles of each reactant by their respective stoichiometric coefficients from the balanced equation: moles of Cl2 / 1 = 0.282 moles of F2 / 3 = 0.088 The smaller value indicates that F2 is the limiting reactant.

Step 3: Calculate the theoretical yield

With the limiting reactant known, we can now find the theoretical yield (in grams) of the product ClF3. Using the stoichiometry from the balanced equation, calculate the moles of ClF3 that should form from the given amount of limiting reactant (0.263 mol of F2): moles of ClF3 = 2 * moles of F2 = 2 * 0.263 = 0.526 mol Now, convert the moles of ClF3 into grams, using the molar mass of ClF3 (i.e., 35.45 g/mol Cl + 3 * 19.00 g/mol = 92.45 g/mol): theoretical yield of ClF3 = (0.526 mol) * (92.45 g/mol) = 48.61 g

Step 4: Calculate the excess reactant left over

Since chlorine is the excess reactant, we need to determine how many grams of chlorine are left over after the reaction. First, find the amount of chlorine that reacted with the fluorine: moles of reacted Cl2 = 0.263 mol F2 / 3 * 2 = 0.175 mol Now calculate the grams of leftover Cl2: grams of leftover Cl2 = (0.282 mol - 0.175 mol) * 35.45 g/mol = 3.79 g

Step 5: Calculate the percent yield

Finally, we calculate the percent yield of the reaction. We are given that the actual yield is 12.50 g of ClF3. Use the equation: percent yield = (actual yield / theoretical yield) * 100% percent yield = (12.50 g / 48.61 g) * 100% = 25.72% So, the percent yield for this reaction is approximately 25.72%.

Key Concepts

Balanced Chemical Equation

In any chemical reaction, creating a balanced chemical equation is a crucial first step. It represents the symbolic translation of the reaction, showing the reactants turning into products in a way that obeys the law of conservation of mass. In the case of chlorine ( Cl_{2} ) reacting with fluorine ( F_{2} ) to produce chlorine trifluoride (ClF3), initially, the unbalanced equation may look like:

Cl2 + F2 → ClF3.

Balancing this equation requires ensuring that the number of atoms of each element is the same on both sides of the equation. By adjusting coefficients, you end up with:

Cl2 + 3F2 → 2ClF3.

This balanced equation indicates that one molecule of chlorine reacts with three molecules of fluorine to produce two molecules of ClF3. Every balanced equation must reflect this kind of stoichiometric relationship to accurately represent the reaction.

Percent Yield

The percent yield is a measure used in chemistry to assess the efficiency of a chemical reaction. It compares the actual yield, which is the amount of product obtained from an experiment, with the theoretical yield, which is the maximum possible amount of product predicted by stoichiometry.

• Formula: Percent yield = (actual yield / theoretical yield) × 100%

For the reaction of chlorine and fluorine forming ClF3, if the actual recovered mass is 12.50 g and the theoretical yield is calculated to be 48.61 g, the percent yield is calculated using:

Percent yield = (12.50 / 48.61) × 100% ≈ 25.72%.

This percentage reveals how close the reaction's actual output was to its potential, highlighting the discrepancies due to factors like unoptimized reaction conditions, side reactions, or experimental errors.

Theoretical Yield

The theoretical yield is the calculated maximum amount of product that can form in an ideal chemical reaction, based on the amount of limiting reactant. Start by identifying the limiting reactant—this is the reactant that will be completely consumed first, thus determining the maximum possible product.

In our example, fluorine ( F_{2} ) is the limiting reactant. Calculations begin with the amount of limiting reactant and proceed using the stoichiometry from the balanced equation. Here, from 0.263 moles of F_{2} , you can predict double that amount in moles of ClF_{3} due to the equation's 1:2 stoichiometry:

moles of ClF3 = 2 × 0.263 = 0.526

Convert moles of ClF_{3} to grams using its molar mass (92.45 g/mol):

Theoretical yield = 0.526 mol × 92.45 g/mol = 48.61 g of ClF_{3} .

This calculation represents the predicted amount of product under perfect conditions.

Excess Reactant

After determining the limiting reactant, the excess reactant is the substance that remains after the reaction is complete. In this reaction, chlorine ( Cl_{2} ) is the excess reactant.

To find how much Cl_{2} is left unreacted, calculate the amount of Cl_{2} that reacted with F_{2} .

Using stoichiometry:

• For every 3 moles of F_{2} , 1 mole of Cl_{2} reacts, so for 0.263 moles of F_{2} , the moles of reacted Cl_{2} = 0.263 / 3 × 2 = 0.175.
• Initially, there were 0.282 moles of Cl_{2} .

Subtract to find the leftover amount in moles:

Moles of leftover Cl_{2} = 0.282 - 0.175 = 0.107

Convert this remaining Cl_{2} into grams using its molar mass (70.90 g/mol):

Grams of leftover Cl_{2} = 0.107 mol × 70.90 g/mol = 7.58 g.

This quantity of excess reactant is present after the reaction has proceeded to completion.