a)
nH2 = 3.36/22.4 = 0.15 (mol)
Fe + H2SO4 => FeSO4 + H2
0.15........................................0.15
mFe = 0.15*56 = 8.4 (g)
b)
nCu = 6.4/64 = 0.1 (mol)
Cu + 2H2SO4 (đ) => CuSO4 + SO2 + 2H2O
0.1...............................................0.1
VSO2 = 0.1*22.4 = 2.24(l)