Question

Answer 1

a) $n_{H_2} = \dfrac{3,09875}{24,79} = 0,125(mol)$

$2Al + 6HCl \to 2AlCl_3 + 3H_2$
$n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{0,25}{3}(mol)$
$m_{Al\ pư} = \dfrac{0,25}{3}.27 = 2,25(gam)$

b) $n_{HCl} = 2n_{H_2} = 0,25(mol)$

$C\%_{HCl} = \dfrac{0,25.36,5}{200}.100\% = 4,5625\%$

c) $m_{dd\ sau\ pư } = m_{Al} + m_{dd\ HCl} - m_{H_2} = 2,25 + 200 - 0,125.2 =202(gam)$

$C\%_{AlCl_3} = \dfrac{\dfrac{0,25}{3}.133,5}{202}.100\% = 5,5\%$