Question

A sample of 6.33 g of solid calcium hydroxide is added to 28.0 mL of 0.500 M aqueous hydrochloric acid. Enter the balanced chemical equation for the reaction.

What are 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

Given :

Mass of Ca(OH)2 =6.33 g

Volume of HCl = 28.0 mL = 0.028 L

[HCl]= 0.500 M

Solution:

Reaction:

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

Limiting reactant :

Calculation of moles of Ca(OH)2

Mol Ca(OH)2 = Mass in g / Molar mass

= 6.33 g x 1 mol Ca(OH)2 / 74.093 g

= 0.0854 mol

Calculation of moles of HCl

Mol HCl = Volume in L x Molarity of HCl

= 0.028 L x 0.500 M

= 0.014 mol HCl

Calculation of moles of Salt :

Moles of salt from Ca(OH)2 :

Mol of CaCl2 = mol Ca(OH)2 x 1 mol CaCl2 / 1 mol Ca(OH)2

= 0.0854 mol CaCl2

Moles of salt from mole HCl

Moles of CaCl2 = mol HCl x 1 mol CaCl2 / 2 mol HCl

= 0.007 mol CaCl2

Moles of product (CaCl2 ) from HCl are less so HCl is limiting reactant :

Calculation of grams of CaCl2 :

Maximum moles of CaCl2 are formed from limiting reactant so we use it.

Moles of CaCl2 = 0.007

Mass of CaCl2 = Moles of CaCl2 x molar mass

= 0.007 moles x 110.98 g per mol

= 0.777 g

Calculation of Excess reactants :

Ca(OH)2 is excess reactant

Number of mole of excess reactant required to react with limiting reactant

= 0.014 mole so HCl x 1 mol Ca(OH)2 / 2 mol HCl

= 0.007 moles of Ca(OH)2

Number of excess moles of Ca(OH)2 = 0.0854 – 0.007 = 0.078 mol

Mass of Ca(OH)2 = Moles x molar mass

= 0.078 mol x 74.093 g per mol

= 5.81 g

Mass of Mass of excess reactant = 5.81 g