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 11.0 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

Balanced Chemical equation is:

CaCO3(s) + 2HCl --> CaCl2(aq) + CO2(g) + H2O

no. of moles of CaCO3 = 25.0 / 100.09 = 0.25 moles

no. of moles of HCl = 11.0/36.5 = 0.30 moles

for each mole of CaCO3, HCl needed = 2 mole

For, 0.25 moles of CaCO3, HCl needed = 2 x 0.25 = 0.5 mole

Since, HCl present is less. The limiting reagent is HCl.

So, 0.30 mole of HCl will react with 0.15 mol of CaCO3

no. of moles of CaCl2 produced = 0.15 mole

weight of CaCl2 formd = 0.15 x 111 = 16.65 g of CaCl2 formed

Excess CaCO3 remain = 0.25 - 0.15 = 0.1 mole = 0.1 x 100.09 = 10 g of caCO3 will remain