Fe + CuSO4 ----> FeSO4 + Cu
a______________________a
n fe= 25/14 mol
n Fe dư = 25/14 - a mol
=> (25/14-a)*56 + 64a = 101.3
=>a=0.1265 mol
=> m Fe = 9.1 g
=> V cuso4 = 0.1265/0.1=1.265 l
c. CuSO4 + 2NaOH ---> Cu(OH)2 + Na2SO4
=> n cuso4 . 5H2O = 12.5/250=0.05 mol
m NaOH = 40*15%=6 g
n NaOH = 6/40=0.15 mol
=> Cuso4 het , NaOh dư
=> n Cu(OH)2 = 0.05 mol
=> m cu(OH)2 = 98*0.05=4.9 g
=> m Na2SO4=0.05*142=7.1 g
=> m NaOh dư= (0.15-0.1)*40=2g
=> m dd = 12.5+ 40-4.9=47. 6 g
=> C% Na2SO4=....
=> C%NaOH=.....