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Chapter 9: Problem 84

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?

Short Answer

Expert verified
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 by step solution

01

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
02

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.
03

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
04

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
05

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%.

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