A sample of 9.60 g of solid calcium hydroxide is added to 34.5 mL of 0.320 M aqueous hydrochloric acid. Write the balanced chemical equation for the reaction. .

Question

How many grams of salt are formed after the reaction is complete? How many grams of the excess reactant remain after the reaction is complete?

Answer 1

This is an acid-base/neutralization reaction. The balanced equation is...

Ca(OH)₂(s) + 2HCl(aq) ==> CaCl₂(aq) + 2H₂O(l)

Limiting reactant:

For Ca(OH)₂: 9.60 g Ca(OH)₂ x 1 mol/74.09 g x 1 mol CaCl₂/mol Ca(OH)₂ x 111 g/mol = 14.38 g CaCl₂ formed

For HCl: 34.5 ml x 1L/1000 ml x 0.320 mol/L x 1 mol CaCl₂/2 mol HCl x 111 g/mol = 0.612 g CaCl₂ formed

LIMITING reactant = HCl

Grams salt formed = 0.612 g CaCl₂ formed

Grams excess reactant (calcium hydroxide) remaining:

34.5 ml x 1L/1000 ml x 0.320 mol/L = 0.0110 moles HCl

moles Ca(OH)₂ used = 0.0110 mol HCl x 1 mol Ca(OH)₂/2 mol HCl = 0.00552 mole Ca(OH)₂ used

How many moles of Ca(OH)₂ did we start with? 9.60 g Ca(OH)₂ x 1 mol/74.09 g = 0.1296 moles

Grams Ca(OH)₂ remaining = 0.1296 mol - 0.00552 mol x 74.09 g/mol = 9.19 g remaining