The first step is to write the full equation:
#CuCl_2 + 2NaOH -> Cu(OH)_2 + 2NaCl#
Using molar quantities of each compound, we can establish that 1 Mole of copper(II) chloride will give 1 Mole of copper(II) hydroxide, when fully reacted.
The amount (in grams) of #CuCl_2# we are using can be calculated by using the concentration and the volume.
#("56 mL") / ("1000 mL") = 0.056 * ("0.779 M" * M_r " of " CuCl_2)#
The #M_r# of #CuCl_2# is #"134.5 g mol"^-1#, so the total amount of available copper(II) chloride is therefore #"5.867 g"#.
Using the 1:1 molar ratio of #CuCl_2 : Cu(OH)_2#, we can work out the final amount of #Cu(OH)_2# as the number of moles of copper(II) chloride #xx# molecular mass of copper(II) hydroxide.
#("5.867 g")/("134.5 gmol"^-1) * M_r " of " Cu(OH)_2 = "4.25 g"# of #Cu(OH)_2#
