2Al + 6HCl => 2AlCl3 + 3H2
0.8 2.4 0.8 1.2 (mol)
nAl = m/M= 21.6/27 = 0.8 (mol)
V = VH2= nx22,4= 1.2 x 22.4 = 26.88 (l)
mHCl = n.M = 2.4 x 36.5 = 87.6 (g)
=> mdd HCl = 87.6x100/7.3 = 1200 (g)
mAlCl3 = n.M = 133.5x0.8 = 106.8 (g)
mdd sau pứ = 1200 + 21.6 - 2.4 = 1219.2 (g)
C% = 106.8x100/1219.2 = 8.76 %