A 15.00 g sample of calcium carbonate was placed in excess hydrochloric acid (HCl). A reaction occurred producing calcium chloride, water, and carbon dioxide.

Question

Formula: CaCO3 + 2HCl = CaCl2 + H2O + CO2 (already balanced)

 

• How many moles of CaCO3 reacted?
• Theoretically, how many grams water should be produced?

 

Answer both questions, show work please

Answer 1

Convert to moles using molecular weight. 

 

Molecular weight CaCO₃ = 1(40 g/mol Ca) + 1(12 g/mol C) + 3(16 g/mol O) = 100 g/mol CaCo₃

 

15.00 g CaCO₃ ÷ 100 g/mol CaCO₃ = 0.15 mol CaCO₃

 

CaCO₃ is the limiting reagent since HCl is in excess. That also means all 0.15 mol CaCO₃ reacted. We use the limiting reagent mole CaCO₃ to calculate mole of H₂O produced.

Stoichiometry is 1:1 ; CaCO₃ to H₂O (so the number moles will be the same!)

 

0.15 mol CaCO₃ × (1 H₂O / 1 CaCO₃) = 0.15 mol H₂O

 

Molecular weight of H₂O = 2(1 g/mol H) + 1(16 g/mol O) = 18 g/mol H₂O

 

0.15 mol H₂O × 18 g/mol H₂O = 2.7 g H₂O