\(a) n_{HCl} = \dfrac{150.7,3\%}{36,5} = 0,3(mol)\\ 2Al + 6HCl \to 2AlCl_3 + 3H_2\\ n_{Al} = \dfrac{1}{3}n_{HCl} = 0,1(mol)\\ m = 0,1.27 = 2,7(gam)\\ b) n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,15(mol)\\ V_{H_2} = 0,15.22,4= 3,36(lít)\\ c) m_{dd\ sau\ pư} = 2,7 + 150 -0,15.2 = 152,4(gam)\\ n_{AlCl_3} =n_{Al} = 0,1(mol)\\ C\%_{AlCl_3} = \dfrac{0,1.133,5}{152,4}.100\% = 8,76\%\)
\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\)
\(m_{HCl}=10,995g\)
=> nHCl = 0,3 mol
a, - THeo PTHH : nAl = 0,1 mol
=> mAl = 2,7g
b, Theo PTHH : nH2 = 0,45 mol
=> Vh2 = 10,08l
c, Theo PTHH : nAlCl3 = 0,3 mol
=> mAlCl3 = 40,05 g
=> C% = 26,22%
