Question

A sample of 9.02 g of solid calcium hydroxide is added to 35.5 mL of 0.260 M aqueous hydrochloric acid. Enter the balanced chemical equation for the reaction. Physical states are optional and not graded.

A) What is the limiting reactant?

B) How many grams of salt is formed after the reaction is complete?

C) How many grams of the excess reaction remain after the reaction is complete?

Answer 1

Ca(OH)2 + 2HCl = CaCl2 + 2H2O

mol of base = mass/MW = 9.02/74.0927 = 0.12173

mol of acid = MV = (35.5*10^-3)*0.260 = 0.00923

ratio is 1:2 so

0.00923 mol of acid --> 1/2*0.00923 = 0.004615 mol of base

a)

Clearly, HCl is limiting, since it will react and limit the reaction

b)

mass of CaCl2 formed from

0.004615 mol of HCl --> 1/2*0.004615 = 0.004615 mol of CaCl2

mass = mol*MW = 0.004615 *110.978 = 0.5121 g of CaCl2

c)

excess leftover

mol of Ca(OH)2 left = 0.12173 - 0.004615 = 0.117115 mol of Ca(OH)2

mass = mol*MW = 0.117115*74.0927 = 8.67 g of Ca(OH)2 left