\(n_{Al}=\dfrac{2,7}{27}=0,1\left(mol\right)\\
pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1 0,3 0,1 0,15
\(m_{HCl}=0,3.36,5=10,95g\\
m_{\text{dd}HCl}=\dfrac{10,95.100}{9,8}=111,73g\)
\(m_{\text{dd}\left(AlCl_3\right)}=2,7+111,73-\left(0,15.2\right)=114,13g\\
C\%=\dfrac{0,1.133,5}{114,13}.100\%=11,7\%\)
