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                           Â
1mol
0.001Â Â Â Â Â Â Â Â Â
0.001Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
??
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
