Question

How many grams of calcium chloride will be produced when 25.0 g of calcium carbonate are combined with 11.0 g of hydrochloric acid?

Answer 1

The equation is as follows

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

1 eq of CaCO3 require 2 eq of hydrochloric acid to produce 1 equivalant of CaCl2

Molecular weights of

CaCO₃-100.08 g/mol

HCl- 36.46 g/mol

CaCl₂- 110.98 g/mol

25 /100.08 = 0.2498 moles of CaCO3 , and 2 x 0.2498 = 0.499 x 36.46= 18.21 g of HCL require but HCL not that much available to react with all CaCO3

11 /36.46= 0.3017 mole of HCl will react with 0.5 x 0.3017= 0 .150 x 100.08=15.09 g of CaCO3 will be used in the reaction.

Hence

100.08 g CaCO3 gives 110.98 g of CaCl2

so

15.09 g of CaCO3 gives = 16.58 gm of CaCl2 will gert produced in the reaction.