\(n_{H2}=\frac{3,36}{22,4}=0,15\left(mol\right)\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
0,1_____________________________0,15
\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)
\(m_{Al2O3}=3,72-0,1.27=1,02\left(g\right)\)
\(\rightarrow n_{AlCl3}=n_{Al}+2n_{Al2O3}=0,12\left(mol\right)\)
\(\rightarrow n_{Al2O3}=\frac{1,02}{102}=0,1\left(mol\right)\)
Khối lượng dd sau phản ứng
\(m=3,71+131,4-0,15.2=134,82\)
\(\rightarrow C\%_{AlCl3}=\frac{0,12.133,5}{134,82}=11,88\%\)
Vdd sau phản ứng \(\frac{131,1}{1,04}=126\left(ml\right)=0,126l\)
\(\rightarrow CM_{AlCl3}=\frac{0,12}{0,126}=0,95M\)