\(n_{HCl}=\dfrac{300.7,3}{36,5.100}=0,6mol\)
\(n_{Al}=\dfrac{2,7}{27}=0,1mol\)
2Al+6HCl\(\rightarrow\)2AlCl3+3H2
\(\dfrac{0,1}{2}< \dfrac{0,6}{6}\) suy ra: Al hết, HCl dư
nHCl=3nAl=0,3mol
nHCl dư=0,6-0,3=0,3mol
mHCldư=0,3.36,5=10,95g
\(n_{H_2}=\dfrac{3}{2}.n_{Al}=0,15mol\)
\(V_{H_2}=0,15.22,4=3,36l\)