If you mix 0.91 L of 0.412 M CuCl2 with 1.92 L of 0.547 M NaOH, how many grams of Cu(OH)2 will you make?

Question

CuCl₂ + NaOH --> Cu(OH)₂ + NaCl

Answer 1

Find number of moles for each reactant

no of moles = volume/molarity

no of moles of CuCl2 = 0.91 L/0.412 M

= 2.21 moles

no of moles of NaOH = 1.92 L/0.547 M

= 3.51 moles

From the stoichiometric ratio, 1 mole of CuCl2 will need 2 moles of NaOH, so 2.21 moles of CuCl2 needs 4.42 moles NaOH but we have 3.51 moles NaOH, therefore, NaOH is the limiting reactant and there will be excess CuCl2 left.

3.51 moles of NaOH will require 3.51/2 moles of 1.76 moles of CuCl2.

Since 1 mole of CuCl2 will produce 1 mole of Cu(OH)2, 1.76 moles of CuCl2 will produce 1.76 moles of Cu(OH)2.

To find the mass of Cu(OH)2 produced, we use

no. of moles of Cu(OH)2 = mass of Cu(OH)2 produced/molar mass of Cu(OH)2

1.76 moles Cu(OH)2 = mass of Cu(OH)2 produced/97.56 g/mol

171.22 g Cu(OH)2 = mass of Cu(OH)2 produced