What volume (in mL) of 4.46 M HCl is required to fully react with 19.3 mL of 2.13 M Ca(OH)2?

Question

Consider the following balanced equation.

2HCl (aq) + Ca(OH)₂ (aq)  2H₂O (l) + CaCl₂ (aq)

Answer 1

2HCl (aq) + Ca(OH)₂ (aq)  2H₂O (l) + CaCl₂ (aq) ... balanced equation

Stoichiometry problem. It takes 2 moles HCl for each 1 mole Ca(OH)₂.

moles Ca(OH)₂ present = 19.3 ml x 1 L/1000 ml x 2.13 mol/L = 0.041109 moles

moles HCl needed = 0.041109 mol Ca(OH)₂ x 2 mol HCl/mol Ca(OH)₂ = 0.082218 mol HCl

Volume of HCl:

(x L)(4.46 mol/L) = 0.082218 mols

x = 0.01843 L x 1000 ml/L = 18.4 mls (to 3 significant figures)