\(n_{HCl}=0,2.1=0,2\left(mol\right)\)
\(m_{Al}=12,9-2,9=10\left(g\right)\)
\(n_{Al}=\dfrac{10}{27}\approx0,37\left(mol\right)\)
PTHH :
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,066 0,2
\(\dfrac{0,37}{2}>\dfrac{0,2}{6}\) --> Tính theo HCl
\(m_{Alpu}=0,66.27=1,8\left(g\right)\)