nNaOH=0.4mol
Al2O3 + 2NaOH -> 2NaAlO2 + H20
=> nAl2O3=0.2 mol
tinh nH2SO4=0.79 mol
viet PTHH cua 3 oxit voi H2SO4 , thay nAl2O3=0.2(mol)
dat x= n MgO; y=nFe2O3
ta co he phuong trinh
40x + 160y= 30 - 0.2 X 102
x + 3y= 0.79-0.2 X 3
giai duoc x = 0.04; y= 0.05(mol)
Tu do ta tinh % moi oxit...