Question

A sample of 9.20 g of solid calcium hydroxide is added to 37.5 mL of 0.440 M aqueous hydrochloric acid. Enter the balanced chemical equation for the reaction. Physical states are optional and not graded. What is the limiting reactant? 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

Number of moles of HCl , n = Molarity x volume in L

= 0.440 M x 37.5 mL x 10^-3 L/mL

= 0.0165 moles

Mass of HCl , m = Number of moles x molar mass

= 0.0165 mol x 36.5 (g/mol)

= 0.602 g

The balanced equation is

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

Molar mass 74 36.5 111 (g / mol)

From this reaction ,

74 g of Ca(OH)₂ reacts with 2x36.5=71 g of HCl

M g of Ca(OH)₂ reacts with 0.602 g of HCl

M = ( 0.602x74)/71

= 0.627 g of Ca(OH)₂

So 9.20 - 0.627 = 8.573 g of Ca(OH)₂ left unreacted so it is the excess reactant

Since all the mass of HCl completly reacted it is the limiting reactant.

Again from the reaction,

2x36.5=71 g of HCl produces 111 g of CaCl₂

0.602 g of HCl produces N g of CaCl₂

N = ( 0.602x111) / 71

= 0.941 g