A sample of 6.71g of solid calcium hydroxide is added to 30.5 mL of 0.430M aqueous hydrochloric acid. Write the balanced chemical equation for the reaction. Physical states are optional.

Question

What is the limiting reactant?

calcium hydroxide

or

hydrochloric acid

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: 6.71 g Ca(OH)₂ x 1 mol/74.09 g x 1 mol CaCl₂/mol Ca(OH)₂ x 111 g/mol = 10.1 g CaCl₂ formed

For HCl: 30.5 ml x 1L/1000 ml x 0.430 mol/L x 1 mol CaCl₂/2 mol HCl x 111 g/mol = 0.728 g CaCl₂ formed

LIMITING reactant = HCl

Grams salt formed = 0.728 g CaCl₂ formed

Grams excess reactant (calcium hydroxide) remaining:

30.5 ml x 1L/1000 ml x 0.430 mol/L = 0.0131 moles HCl

moles Ca(OH)₂ used = 0.0131 mol HCl x 1 mol Ca(OH)₂/2 mol HCl = 0.00656 mole Ca(OH)₂ used

How many moles of Ca(OH)₂ did we start with? 6.71 g Ca(OH)₂ x 1 mol/74.09 g = 0.0906 moles

Grams Ca(OH)₂ remaining = 0.0906 mol - 0.00656 mol x 74.09 g/mol = 6.23 g remaining