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: Determine the total volume of CO_2 produced. From the graph, the reaction for Experiment A stops at a maximum volume of CO_2 gas. Volume of CO_2 = 460 cm^3 Step 2: Convert the volume of CO_2 to dm^3. Since the molar gas volume is given in dm^3·mol^-1, convert the volume of CO_2 from cm^3 to dm^3. 1 dm^3 = 1000 cm^3 Volume of CO_2 = (460)/(1000) dm^3 = 0.460 dm^3 Step 3: Calculate the moles of CO_2 produced. Use the given molar gas volume at 40°C, which is 25.7 dm^3·mol^-1. Moles of CO_2 = Volume of CO_2Molar gas volume Moles of CO_2 = 0.460 dm^325.7 dm^3·mol^-1 Moles of CO_2 ≈ 0.0178988 mol Step 4: Determine the moles of CaCO_3 used. From the balanced chemical equation: CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + H_2O(l) + CO_2(g) The mole ratio of CaCO_3 to CO_2 is 1:1. Therefore, moles of CaCO_3 used = moles of CO_2 produced. Moles of CaCO_3 = 0.0178988 mol Step 5: Calculate the molar mass of CaCO_3. Molar mass of Ca = 40.08 g/mol Molar mass of C = 12.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of CaCO_3 = 40.08 + 12.01 + (3 × 16.00) Molar mass of CaCO_3 = 40.08 + 12.01 + 48.00 = 100.09 g/mol Step 6: Calculate the mass of CaCO_3 used. Mass of CaCO_3 = Moles of CaCO_3 × Molar mass of CaCO_3 Mass of CaCO_3 = 0.0178988 mol × 100.09 g/mol Mass of CaCO_3 ≈ 1.7915 g Rounding to two decimal places: Mass of CaCO_3 ≈ 1.79 g The mass of CaCO_3(s) used in experiment A is 1.79 g. That's 2 down. 3 left today — send the next one.