Câu 1 :
$a) 2Al + 6HCl \to 2AlCl_3 + 3H_2$
$b) n_{Al} = \dfrac{0,54}{27} = 0,02(mol)$
Theo PTHH, $n_{H_2} = \dfrac{3}{2}n_{Al} = 0,03(mol)$
$V_{H_2} = 0,03.22,4 = 0,672(lít)$
$n_{HCl} = 2n_{H_2} = 0,06(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,06.36,5}{10\%} = 21,9(gam)$
$c) m_{dd\ sau\ pư} = 0,54 + 21,9 -0,03.2 = 22,38(gam)$
$C\%_{AlCl_3} = \dfrac{0,02.133,5}{22,38}.100\% = 11,93\%$
Câu 2 :
$a) Mg + 2HCl \to MgCl_2 + H_2$
$b)
$n_{Mg} = \dfrac{0,48}{24} = 0,02(mol)$
Theo PTHH, $n_{H_2} =n_{Mg} = 0,02(mol)$
$V_{H_2} = 0,02.22,4 = 0,448(lít)$
$n_{HCl} = 2n_{H_2} = 0,04(mol) \Rightarrow m_{dd\ HCl} = \dfrac{0,04.36,5}{10\%} = 14,6(gam)$
$c) m_{dd\ sau\ pư} = 0,48 + 14,6 -0,02.2 = 15,04(gam)$
$C\%_{MgCl_2} = \dfrac{0,02.95}{15,04}.100\% = 12,63\%$
