Question

Reaction 2: 10 mL of 0.1 M CuSO4 and 10 mL of 0.1 M NaOH Calculate the theoretical yield in grams of the product

Answer 1

moles of CuSO4 = 10 x 0.1 / 1000 = 0.001

moles of NaOH = 10 x 0.1 / 1000 = 0.001

CuSO4 + 2 NaOH ---------------> Cu(OH)2 (s) + Na2SO4

1mol          2 mol                             1mol

0.001          0.001                              ??

limiting reagent NaOH

2 mol NaOH ----------------> 1mol Cu(OH)2 product

0.001 mol NaoH -----------> 0.001 /2 = 5 x 10^-4 moles

Cu(OH)2 molar mass = 74 .09 g/mol

mass = moles x molar mass

          = 5 x 10^-4 x 74.09

          = 0.037 g

theoretical yield of the product = 0.037 g