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 are the solutions to the problems you've provided. Problem 3.92: Overall Percent Yield For successive reactions, the overall percent yield is the product of the individual percent yields (expressed as decimals). Step 1: Convert individual percent yields to decimals. Yield of D E = 48\% = 0.48 Yield of E F = 73\% = 0.73 Step 2: Calculate the overall yield. Overall Yield = Yield (D E) × Yield (E F) Overall Yield = 0.48 × 0.73 = 0.3504 Step 3: Convert the overall yield back to a percentage. Overall Percent Yield = 0.3504 × 100\% = 35.04\% Rounding to two significant figures: 35% Problem 3.93: Percent Yield of Tungsten(VI) Oxide Reaction The balanced chemical equation for the reaction is: WO_3(s) + 3H_2(g) W(s) + 3H_2O(l) Given: Mass of WO_3 = 45.5 g Actual yield of H_2O = 9.60 mL Density of H_2O = 1.00 g/mL Step 1: Calculate the molar masses of WO_3 and H_2O. Molar mass of WO_3 = 183.84 g/mol (W) + (3 × 16.00 g/mol (O)) = 231.84 g/mol Molar mass of H_2O = (2 × 1.008 g/mol (H)) + 16.00 g/mol (O) = 18.016 g/mol Step 2: Convert the actual yield of water from volume to mass. Actual mass of H_2O = volume × density = 9.60 mL × 1.00 g/mL = 9.60 g Step 3: Calculate the moles of WO_3. moles of WO_3 = 45.5 g231.84 g/mol = 0.19625 mol Step 4: Calculate the theoretical moles of H_2O produced. From the balanced equation, 1 mol of WO_3 produces 3 mol of H_2O. theoretical moles of H_2O = 0.19625 mol WO_3 × 3 mol H_2O1 mol WO_3 = 0.58875 mol H_2O Step 5: Calculate the theoretical mass (yield) of H_2O. theoretical mass of H_2O = 0.58875 mol × 18.016 g/mol = 10.607 g Step 6: Calculate the percent yield. Percent Yield = Actual YieldTheoretical Yield × 100\% Percent Yield = 9.60 g10.607 g × 100\% = 90.506\% Rounding to three significant figures: 90.5% Problem 3.94: Percent Yield of Phosphorus Trichloride Reaction The balanced chemical equation for the reaction is: PCl_3(l) + 3H_2O(l) 3HCl(aq) + H_3PO_3(aq) Given: Mass of PCl_3 = 200. g Actual yield of HCl = 128 g Water is in excess. Step 1: Calculate the molar masses of PCl_3 and HCl. Molar mass of PCl_3 = 30.97 g/mol (P) + (3 × 35.45 g/mol (Cl)) = 137.32 g/mol Molar mass of HCl = 1.008 g/mol (H) + 35.45 g/mol (Cl) = 36.458 g/mol Step 2: Calculate the moles of PCl_3. moles of PCl_3 = 200. g137.32 g/mol = 1.45645 mol Step 3: Calculate the theoretical moles of HCl produced. From the balanced equation, 1 mol of PCl_3 produces 3 mol of HCl. theoretical moles of HCl = 1.45645 mol PCl_3 × 3 mol HCl1 mol PCl_3 = 4.36935 mol HCl Step 4: Calculate the theoretical mass (yield) of HCl. theoretical mass of HCl = 4.36935 mol × 36.458 g/mol = 159.29 g Step 5: Calculate the percent yield. Percent Yield = Actual YieldTheoretical Yield × 100\% Percent Yield = 128 g159.29 g × 100\% = 80.356\% Rounding to three significant figures: 80.4% 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. Since 0.63469 mol of Cl_2 is less than 1.2779 mol of CH_4, 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 Problem 3.96: Mass of Calcium Nitride Produced The balanced chemical equation for the reaction is: 3Ca(s) + N_2(g) Ca_3N_2(s) Given: Mass of Ca = 56.8 g Mass of N_2 = 30.5 g Percent yield = 93.0\% Step 1: Calculate the molar masses of Ca, N_2, and Ca_3N_2. Molar mass of Ca = 40.08 g/mol Molar mass of N_2 = 2 × 14.01 g/mol (N) = 28.02 g/mol Molar mass of Ca_3N_2 = (3 × 40.08 g/mol (Ca)) + (2 × 14.01 g/mol (N)) = 120.24 + 28.02 = 148.26 g/mol Step 2: Calculate the moles of each reactant. moles of Ca = 56.8 g40.08 g/mol = 1.4172 mol moles of N_2 = 30.5 g28.02 g/mol = 1.0885 mol Step 3: Determine the limiting reactant. From the balanced equation, 3 mol of Ca reacts with 1 mol of N_2. To react all 1.4172 mol of Ca, we need: 1.4172 mol Ca × 1 mol N_23 mol Ca = 0.4724 mol N_2 Since we have 1.0885 mol of N_2 (which is more than 0.4724 mol needed), calcium (Ca) is the limiting reactant. Step 4: Calculate the theoretical moles of Ca_3N_2 produced from the limiting reactant. From the balanced equation, 3 mol of Ca produces 1 mol of Ca_3N_2. theoretical moles of Ca_3N_2 = 1.4172 mol Ca × 1 mol Ca_3N_23 mol Ca = 0.4724 mol Ca_3N_2 Step 5: Calculate the theoretical mass (yield) of Ca_3N_2. theoretical mass of Ca_3N_2 = 0.4724 mol × 148.26 g/mol = 70.03 g Step 6: Calculate the actual mass of Ca_3N_2 using the given percent yield. Actual Yield = Theoretical Yield × Percent Yield100\% Actual Yield = 70.03 g × (93.0\%)/(100\%) = 70.03 g × 0.930 = 65.1279 g Rounding to three significant figures: 65.1 g That's 2 down. 3 left today — send the next one.