P1 + H2SO4 loãng dư, chỉ Fe p/ư
Fe + H2SO4 -> FeSO4 + H2
0.1 0.1
P2 + H2SO4 đặc nóng, cả 2 KL p/ư
2Fe + 6H2SO4 -> Fe2(SO4)3 + 3SO2 + 6H2O
0.1 0.05 0.15
Cu + 2H2SO4 -> CuSO4 + SO2 + 2H2O
0.15 <- 0.15 0.15
%Fe = 0.1 x 56 / (0.1 x 56 + 0.15 x 64) = 36.84%
%Cu = 100% - %Fe = 63.16%
m dd Y = 0.1 x 56 + 0.15 x 64 + 100 - 0.3 x 64 = 96g
=>C%Fe2(SO4)3 = 0.05 x 400 x 100 / 96 = 20.83%
C% CuSO4 = 0.15 x 160 x 100 / 96 = 25%