nCuSO4 = \(\frac{50.1,12.15\%}{160}=0,0525mol\)
Fe + CuSO4 => FeSO4 + Cu
0,02-->0,02------->0,02---->0,02
nFe pư= \(\frac{5,16-5}{64-56}=0,02mol\)
mdd = 5+50.1,12 - 0,02.64 = 59,72 (g)
C% CuSO4 = \(\frac{\left(0,0525-0,02\right).160}{59,72}.100\%=8,7\%\)
C% FeSO4 = \(\frac{0,02.152}{59,72}.100\%=5,09\%\)