\(a,n_{HCl}=0,05.1=0,05\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
\(\dfrac{1}{60}\)<--0,05------->\(\dfrac{1}{60}\)
b, \(m_{Al}=\dfrac{1}{60}.27=0,45\left(g\right)\)
c, \(C_{M\left(AlCl_3\right)}=\dfrac{\dfrac{1}{60}}{0,05}=\dfrac{1}{3}M\)