Question

# When calcium carbonate is added to hydrochloric acid, calcium chloride, carbon dioxide, and water are produced....

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?

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

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