2Al+6HCl->2AlCl3+3H2
0,185-----------0,185---0,2775 mol
n Al=\(\dfrac{5}{27}\)=0,185 mol
m HCl=21,9=>n HCl=0,6 mol
=>Al td hết , HCl dư
=>C% AlCl3=\(\dfrac{0,185.133,5}{5+150-0,2775.2}\).100=15,99%
=>C% HCl dư=\(\dfrac{0,045.36,5}{5+150-0,2775.2}\).100=1%