A chemistry student needs to determine the percentage of calcium carbonate (CaCO3) present in a mixture of calcium carbonate and calcium phosphate. The student has the following information: A 1.00-gram sample of the mixture was dissolved and reacted with excess hydrochloric acid (HCl), producing 0.236 grams of carbon dioxide (CO2) gas. The balanced chemical equation for the reaction is: CaCO3 + 2HCl → CaCl2 + CO2 + H2O. The student also knows that the molecular weight of CaCO3 is 100.09 g/mol and that of Ca3(PO4)2 is 310.18 g/mol. What is the percentage of calcium carbonate in the mixture?

Question

A chemistry student needs to determine the percentage of calcium carbonate (CaCO3

Answer 1

First, we need to determine the moles of CO2 produced in the reaction. To do this, we will use the molecular weight of CO2, which is 44.01 g/mol.Moles of CO2 = mass of CO2 / molecular weight of CO2Moles of CO2 = 0.236 g / 44.01 g/mol = 0.00536 molNow, we will use the stoichiometry of the balanced chemical equation to determine the moles of CaCO3 that reacted.1 mol CaCO3 reacts with 1 mol CO2  from the balanced equation So, moles of CaCO3 = moles of CO2 = 0.00536 molNow, we will convert the moles of CaCO3 back to grams using the molecular weight of CaCO3.Mass of CaCO3 = moles of CaCO3  molecular weight of CaCO3Mass of CaCO3 = 0.00536 mol  100.09 g/mol = 0.536 gNow, we can calculate the percentage of calcium carbonate in the mixture.Percentage of CaCO3 =  mass of CaCO3 / mass of the mixture   100Percentage of CaCO3 =  0.536 g / 1.00 g   100 = 53.6%The percentage of calcium carbonate in the mixture is 53.6%.