nFe= 1.3/56= 0,23(mol) ; nCuSO4= 0.015 x 2= 0.03(mol) ; nNaOH= 0.09 x 1 = 0.09 (mol)
ptpư : Fe+ CuSO4 = FeSO4 + Cu (1)
0,03 0.03 0.03 0.03
a) từ pt (1) ta có mA= mFe dư + mCu = (0.23-0.03) x 56+ 64 x 0.03 = 13,12 g
b) ta có pt :
FeSO4+ NaOH = Fe(OH)2 + Na2SO4 (2)
0.03 0.03 0.03
theo pt (2): nFe(OH)2= 0.03 mol
mFe(OH)2 = 0.03x 90= 2.7 (g)