2Al + 6HCl -> 2AlCl3 + 3H2
0.2 0.6 0.2 0.3
nAl = 5.4/27 = 0.2 mol
theo ptpu: nH2 = 0.3 mol
=> V H2 = 0.3 x 22.4 = 6.72 l
nHCl p/ư = 0.6 mol => nHCl dư = 0.06 mol
mdd sau p/ư = 5.4 +100 - 0.3 x 2 = 104.8g
C% HCl dư = 0.06 x 36.5 x 100 / 104.8 = 2.09%
C% AlCl3 = 0.2 x 133.5 x 100 / 104.8 = 11.92%
AlCl3 + 3NaOH -> Al(OH)3 + 3NaCl
0.2 0.6
=> V NaOH = 0.6 / 1 = 0.6 l = 600 ml