\(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)

PTHH:

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\) 

0,2      0,6           0,2          0,3

\(m_{Al}=0,2.27=5,4\left(g\right)\)

\(C_{M\left(HCl\right)}=\dfrac{0,6}{0,6}=1\left(M\right)\)

\(m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

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