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: Calculate the molar mass of calcium carbonate (CaCO_3). The atomic masses are: Ca = 40.08 g/mol, C = 12.01 g/mol, O = 16.00 g/mol. Molar mass of CaCO_3 = 40.08 + 12.01 + (3 × 16.00) = 40.08 + 12.01 + 48.00 = 100.09 g/mol Step 2: Calculate the initial moles of calcium carbonate. Moles of CaCO_3 (initial) = MassMolar mass = 15.0 g100.09 g/mol ≈ 0.149865 mol Step 3: Determine the moles of calcium carbonate that react with 0.2 moles of hydrochloric acid (HCl). The balanced chemical equation is: CaCO_3(s) + 2HCl(aq) CaCl_2(aq) + CO_2(g) + H_2O(l) From the stoichiometry, 1 mole of CaCO_3 reacts with 2 moles of HCl. Therefore, the moles of CaCO_3 required to react with 0.2 moles of HCl are: Moles of CaCO_3 (reacted) = 0.2 mol HCl × 1 mol CaCO_32 mol HCl = 0.1 mol CaCO_3 Step 4: Identify the limiting reactant and calculate the moles of calcium carbonate remaining. Initial moles of CaCO_3 = 0.149865 mol Moles of CaCO_3 that will react = 0.1 mol Since the initial moles of CaCO_3 (0.149865 mol) are greater than the moles of CaCO_3 that will react (0.1 mol), HCl is the limiting reactant, and some CaCO_3 will remain. Moles of CaCO_3 (remaining) = Moles of CaCO_3 (initial) - Moles of CaCO_3 (reacted) Moles of CaCO_3 (remaining) = 0.149865 mol - 0.1 mol = 0.049865 mol Step 5: Convert the remaining moles of calcium carbonate to mass. Mass of CaCO_3 (remaining) = Moles remaining × Molar mass Mass of CaCO_3 (remaining) = 0.049865 mol × 100.09 g/mol ≈ 4.991 g Rounding to three significant figures: Mass of CaCO_3 (remaining) ≈ 4.99 g The amount of calcium carbonate that would remain is 4.99 g.