nAl=\(\frac{4,05}{27}\text{ = 0,15 mol}\)
PTHH:
2Al +6HCl → 2AlCl3 +3H2↑
0,15__0,45_____0,15_____0,225
VH2 = 0,225.22,4= 5,04 (l)
mAlCl3 = 0,15.133,5=20,025 g
C% HCl=\(\frac{0,45.36,5}{500}\text{.100%= 3,285%}\)
C%AlCl3=\(\frac{20,025}{4,05+500-0,225.2}\text{ .100%= 3,97%}\)