2Fe + 6H2SO4đ,n => Fe2(SO4)3 + 3SO2 + 6H2O
Fe2O3 + 3H2SO4đ,n => Fe2(SO4)3 + 3H2O
nSO2 = V/22.4 = 6.72/22.4 = 0.3 (mol)
==> nFe = 0.2 (mol) => mFe = 11.2 (g)
==> mFe2O3 = 27.2 - 11.2 = 16 (g)
nFe2O3 = m/M = 16/160 = 0.1 (mol)
nH2SO4 = 0.9 (mol) => mH2SO4 = n>M = 0.9 x 98 = 88.2 (g)
==> mddH2SO4 = 88.2x100/80 = 110.25 (g)