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.
ScanSolve AI Answer
Step 1: Write the balanced chemical equation for the reaction between methane (CH_4) and chlorine (Cl_2) under UV light. The reaction is a free radical substitution. CH_4(g) + Cl_2(g) UV light CH_3Cl(g) + HCl(g) Step 2: Determine the mole ratio between methane and chlorine from the balanced equation. From the equation, 1 mole of CH_4 reacts with 1 mole of Cl_2. The mole ratio is 1:1. Step 3: Calculate the moles of methane and chlorine provided. Molar mass of CH_4 = 12.01 + 4 × 1.01 = 16.05 g/mol Moles of CH_4 = (32.1 g) / (16.05 g/mol) = 2.00 mol Molar mass of Cl_2 = 2 × 35.45 = 70.90 g/mol Moles of Cl_2 = (70.9 g) / (70.90 g/mol) = 1.00 mol Step 4: Identify the limiting reactant. Since the mole ratio is 1:1, and we have 2.00 mol of CH_4 and 1.00 mol of Cl_2, chlorine (Cl_2) is the limiting reactant. Step 5: Calculate the moles of chloromethane (CH_3Cl) produced based on the limiting reactant. From the balanced equation, 1 mole of Cl_2 produces 1 mole of CH_3Cl. Therefore, 1.00 mol of Cl_2 will produce 1.00 mol of CH_3Cl. Step 6: Calculate the mass of chloromethane produced. Molar mass of CH_3Cl = 12.01 + 3 × 1.01 + 35.45 + 1.01 = 49.49 g/mol Mass of CH_3Cl = (moles of CH_3Cl) × (molar mass of CH_3Cl) Mass of CH_3Cl = (1.00 mol) × (49.49 g/mol) = 49.49 g The mass of chloromethane produced is 49.49 g. What's next? 📸