Question

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

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

A) How many grams of calcium chloride will be produced when 32g of calcium carbonate are combined whith 12.0g of hydrochloric acid?

Mass of CaCl2:

B) Which reactant is in excess and how many grams of this reactant will remain after the reaction is complete?

Mass of excess reactant:

Answer 1

(A) mass of CaCl₂ = 18.3 g

(B) mass of excess reactant = 15.5 g

Explanation

(A) The balanced reaction is : 2 HCl + CaCO₃ CaCl₂ + H₂O + CO₂

Case 1 : Let CaCO₃ be the limiting reactant

mass CaCO₃ = 32 g

moles CaCO₃ = (mass CaCO₃) / (molar mass CaCO₃)

moles CaCO₃ = (32 g) / (100.1 g/mol)

moles CaCO₃ = 0.320 mol

moles CaCl₂ formed = moles CaCO₃ reacted

moles CaCl₂ formed = 0.320 mol

mass CaCl₂ formed = (moles CaCl₂ formed) * (molar mass CaCl₂)

mass CaCl₂ formed = (0.320 mol) * (111 g/mol)

mass CaCl₂ formed = 35.5 g

Case 2 : Let HCl be the limiting reactant

mass HCl = 12.0 g

moles HCl = (mass HCl) / (molar mass HCl)

moles HCl = (12.0 g) / (36.5 g/mol)

moles HCl = 0.329 mol

moles CaCl₂ formed = (moles HCl) / 2

moles CaCl₂ formed = (0.329 mol) / 2

moles CaCl₂ formed = 0.165 mol

mass CaCl₂ formed = (moles CaCl₂ formed) * (molar mass CaCl₂)

mass CaCl₂ formed = (0.165 mol) * (111 g/mol)

mass CaCl₂ formed = 18.3 g

Since less mass of CaCl₂ is formed in Case 2, therefore HCl is the limiting reagent.

mass CaCl₂ formed = 18.3 g

moles CaCl₂ formed = 0.165 mol

moles CaCO₃ reacted = moles CaCl₂ formed

moles CaCO₃ reacted = 0.165 mol

mass CaCO₃ reacted = (moles CaCO₃ reacted) * (molar mass CaCO₃)

mass CaCO₃ reacted = (0.165 mol) * (100.1 g/mol)

mass CaCO₃ reacted = 16.5 g

mass CaCO₃ left = (total mass CaCO₃) - (mass CaCO₃ reacted)

mass CaCO₃ left = (32 g) - (16.5 g)

mass CaCO₃ left = 15.5 g