mAl2(SO4)3 = 200.1,71% = 3,42g
=> nAl(3+) = 2.(3,42/342) = 0,02 mol
Al(3+) + 3OH(-) = Al(OH)3
x.............3x............x mol
Al(3+) + 4OH(-) = Al(OH)4(-)
y.............4y
Ta có : nAl(3+) = x + y = 0,02 mol
Mặt khác : nAl(OH)3 = x = 0,78/78 = 0,01 mol
=> x = y = 0,01 mol
=> nOH(-) = nNaOH = 3x + 4y = 3.0,01 + 4.0,01 = 0,07 mol
=> mNaOH = 0,07.40 = 2,8g