This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here's the solution for problem 3.95. Problem 3.95: Mass of Chloromethane Produced The balanced chemical equation for the reaction is: CH_4(g) + Cl_2(g) CH_3Cl(g) + HCl(g) Given: Mass of CH_4 = 20.5 g Mass of Cl_2 = 45.0 g Percent yield = 75.0\% Step 1: Calculate the molar masses of CH_4, Cl_2, and CH_3Cl. Molar mass of CH_4 = 12.01 g/mol (C) + (4 × 1.008 g/mol (H)) = 16.042 g/mol Molar mass of Cl_2 = 2 × 35.45 g/mol (Cl) = 70.90 g/mol Molar mass of CH_3Cl = 12.01 g/mol (C) + (3 × 1.008 g/mol (H)) + 35.45 g/mol (Cl) = 50.484 g/mol Step 2: Calculate the moles of each reactant. moles of CH_4 = 20.5 g16.042 g/mol = 1.2779 mol moles of Cl_2 = 45.0 g70.90 g/mol = 0.63469 mol Step 3: Determine the limiting reactant. From the balanced equation, 1 mol of CH_4 reacts with 1 mol of Cl_2. To react all 1.2779 mol of CH_4, we would need 1.2779 mol of Cl_2. Since we only have 0.63469 mol of Cl_2, chlorine (Cl_2) is the limiting reactant. Step 4: Calculate the theoretical moles of CH_3Cl produced from the limiting reactant. From the balanced equation, 1 mol of Cl_2 produces 1 mol of CH_3Cl. theoretical moles of CH_3Cl = 0.63469 mol Cl_2 × 1 mol CH_3Cl1 mol Cl_2 = 0.63469 mol CH_3Cl Step 5: Calculate the theoretical mass (yield) of CH_3Cl. theoretical mass of CH_3Cl = 0.63469 mol × 50.484 g/mol = 32.049 g Step 6: Calculate the actual mass of CH_3Cl using the given percent yield. Actual Yield = Theoretical Yield × Percent Yield100\% Actual Yield = 32.049 g × (75.0\%)/(100\%) = 32.049 g × 0.750 = 24.03675 g Rounding to three significant figures: 24.0 g 3 done, 2 left today. You're making progress.