Question

As background, an antacid contains an unknown number of milligrams (mg) of CaCO3 per 2.0 g tablet according to the label. (CaCO3: 100.09 g/mol). Based on the questions below, determine the mg of CaCO3 in the 2.0 g tablet. The reaction by which the antacid neutralizes HCl is 2 HCl(aq) + CaCO3(s) --> CaCl2(aq) + CO2(g) + H2O(l) Q1: How many moles of HCl are neutralized by one mole of CaCO3(s)? ___ mol Q2: 50.0 mL of 0.300 M HCl are added to an Erlenmeyer flask containing a 2.00 g antacid tablet. Calculate the moles of acid in the 50.0 mL of 0.300 M HCl? ___ mol Q3: A student fills her burette with NaOH to the 2.5 mL mark. She titrates her sample of the NaOH until she reaches the endpoint (i.e. all the acid has been neutralized by the NaOH). The volume marking on her burette - at the endpoint - is 52.5 mL. How many mL of NaOH did she use to reach the endpoint? Q4: Some of the 2.0 g of antacid neutralizes the HCl. However, there is excess acid, not neutralized by the antacid. To determine the excess moles of acid, a student used 50.00 mL of 0.100 M NaOH to titrate the excess acid. How many moles of excess acid are there? ___ mol HCl Q5: How many moles of HCl were neutralized by the 2.0 g antacid tablet? ___ mol Q6: How many mg of CaCO3 were in the original 2.00 g tablet? ___ mg

Answer 1