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 25.0 g of calcium carbonate are combined with 14.0 g of hydrochloric acid?Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

I found that there is 21.01 Grams of CaCl2

Answer 1

The balance reaction is as follows:

CaCO3(s) + 2HCl(aq) = CaCl2(aq) + H2O(l) + CO2(g

25.0 g              14.0 g

Now calculate the number of moles of both reactants:

CaCO3: 25.0 g/100.1 g/mole= 0.25 moles
HCl: 14.0 g HCl/36.5 g/mole= moles 0.38 moles

See from the balance reaction; here 2 moles of HCl are required per 1 mole CaCO3, therefore HCl is limiting reactant means HCl reacted completely, and CaCO3 is remain in reaction mixture.

First calculate the number of moles of CaCO3 which are reacted with HCl:

1/2 x 0.38 moles HCl= 0.19 moles CaCO3 reacted

Here total moles of CaCO3 = 0.25 moles

moles of unreacted CaCO3= total moles of CaCO3- moles CaCO3 reacted

= 0.25 moles- 0.19 moles

= 0.06 moles .

Now calculate the grams of this reactant will remain after the reaction is complete is as follows:

0.06 mol CaCO3* 100.1 g/mole= 6.0 g CaCO3.

Thus 6.0 g CaCO3 will remain after the reaction is complete