Question

When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced. How many grams of calcium chloride will be produced when 32.0 g of calcium carbonate are combined with 15.0 g of hydrochloric acid? Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

Answer 1

CaCO3 + 2HCl --------------> CaCl2 + CO2 + H2O

no of moles of CaCO3   = W/G.M.Wt

                                      = 32/100   = 0.32 moles

no of moles of HCl       = W/G.M.Wt

                                     =15/36.5 = 0.41 moles

1 mole of CaCO3 react with 2 moles of HCl

0.32 moles of CaCO3 react with = 2*0.32/1 = 0.64 moles of HCl is required

HCl is limiting reactant

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

0.41 moles of HCl react with CaCO3 to gives = 0.41*1/2    = 0.205 moles of CaCl2

mass of CaCl2 = no of moles * gram molar mass

                          = 0.205*111 = 22.755g of CaCl2

2 moles of HCl react with 1 moles of CaCO3

0.41 moles of HCl react with = 1*0.41/2    = 0.205 moles of CaCO3

CaCO3 is excess reactant

The no of moles of excess reactant will remain after complete reaction = 0.32-0.205   = 0.115 moles of CaCO3

The amount of excess reactant will remains after complete reaction = no of moles * gram molar mass

                                                                                                             = 0.115*100 = 11.5g of CaCO3