Question

CaCO,(s) + 2 HCl(aq) +CaCl, (aq) + H,0(1) + CO,() How many grams of calcium chloride will be produced when 29.0 g of calcium carbonate is combined with 14.0 g of hydrochloric acid? mass of CaCl,: 21.292 Which reactant is in excess? HCI Caco, How many grams of the excess reactant will remain after the reaction is complete? mass of excess reactant: 21

Answer 1

The chemical reaction for the formation of Calcium Chloride (CaCl2) is-

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

i.e from the reaction of 1 mole of CaCO3 and 2 moles of HCl, we get 1 mole of CaCl2

Now given

mass of CaCO3 taken = 29.0 g

So moles of  CaCO3 taken = mass / molar mass of CaCO3

= 29.0 g /100 g/mol

= 0.29 moles

So molees of HCl required fro complete reactio = 0.29 moles * 2

= 0.58 moles

Now given

mass of HCl taken = 14 g

So moles of  HCl taken = mass / molar mass of HCl

= 14 g /36.46 g/mol

= 0.38 moles

That means we have less than required moles of HCl. So here HCl is the limiting reagent.

Thus the final reaction will be as per the limiting reagent. i.e

0.38/2 moles CaCO3 + 0.38 moles HCl ---------> 0.38/2 moles CaCl2 + 0.38/2 moles H2O + 0.38/2 moles CO2

0.19 moles CaCO3 + 0.38 moles HCl ---------> 0.19 moles CaCl2 + 0.19 moles H2O + 0.19 moles CO2

So finally moles of CaCl2 produced = 0.19 moles

So mass of CaCl2 produced = moles * molar mass of CaCl2

= 0.19 moles * 110.98 g/mol

= 21.08 g

Now the excess reagent = CaCO3

Again moles of CaCO3 taken = 0.29 moles

And moles of CaCO3 reacted = 0.19 moles

Then moles of CaCO3 left = 0.29 - 0.19 = 0.10 moles

So mass of CaCO3 left = moles * molar mass CaCO3

= 0.10 moles * 100 g/mol

= 10 g