Question

Consider the following reaction: 2 HCl + CaCO3 → CaCl2 + H2O + CO2

How many grams of calcium chloride can be produced if you begin with 6.19 mL of 1.01 M HCl and 1.700 grams of calcium carbonate?

Answer 1

2HCl + CaCO3 → CaCl2 + H2O + CO2

no of moles of HCl   = molarity * volume in L

                                 = 1.01*0.00619    = 0.0062519 moles

no of moles of CaCO3    = W/G.M.Wt

                                         = 1.7/100   = 0.017 moles

2HCl + CaCO3 → CaCl2 + H2O + CO2

1 mole of CaCO3 react with 2 moles of HCl

0.017 moles of CaCO3 react with= 2*0.017/1    = 0.034 moles of HCl

HCl is limiting reactant

2 moles of HCl react with excess of CaCO3 to gives 1 mole of CaCl2

0.0062519 moles of HCl react with excess of CaCO3 to gives = 1*0.0062519/2    = 0.00312595 moles

mass of CaCl2 = no of moles* gram molar mass

                          = 0.00312595*111    = 0.347g of CaCl2 >>>>answer