`2Al + 6HCl -> 2AlCl_3 + 3H_2 \uparrow`
`0,1` `0,3` `0,1` `0,15` `(mol)`
`n_[HCl]=[5,475.200]/[100.36,5]=0,3(mol)`
`a)m_[Al]=0,1.27=2,7(g)`
`b)V_[H_2]=0,15.22,4=3,36(l)`
`c)C%_[AlCl_3]=[0,1.133,5]/[2,7+200-0,15.2].100=6,6%`
a) $n_{HCl} = \dfrac{100.5,475\%}{36,5} = 0,15(mol)$
$2Al + 6HCl \to 2AlCl_3 + 3H_2$
Theo PTHH : $n_{Al} = \dfrac{1}{3}n_{HCl} = 0,05(mol)$
$m = 0,05.27 = 1,35(gam)$
b) $n_{H_2} = \dfrac{1}{2}n_{HCl} = 0,075(mol)$
$V_{H_2} = 0,075.22,4 = 1,68(lít)$
c) $m_{dd\ sau\ pư} = 1,35 + 200 - 0,075.2 = 201,2(gam)$
$C\%_{AlCl_3} = \dfrac{0,05.133,5}{201,2}.100\% = 3,32\%$