Đáp án + Giải thích các bước giải:

a)

 $m_{H_2SO_4} = \dfrac{m_{dd} . 100\%}{C\%} = \dfrac{300 . 9,8\%}{100\%} = 29,4g$

$\Rightarrow n_{H_2SO_4} = \dfrac{m}{M} = \dfrac{29,4}{2 . 1 + 32 + 4 . 16} = \dfrac{29,4}{98} = 0,3\text{ mol}$

PTHH:

$2Al + 3H_2SO_4 \longrightarrow Al_2(SO_4)_3 + 3H_2$
Theo PTHH:
$n_{H_2} = n_{H_2SO_4} = 0,3\text{ mol}$

$\Rightarrow V_{H_2} = n . 22,4 = 0,3 . 22,4 = 6,72l$
b) Theo PTHH:
$n_{Al} = \dfrac{2}{3}n_{H_2SO_4} = \dfrac{2}{3} . 0,3 = 0,2\text{ mol}$

$\Rightarrow m = m_{Al} = 0,2 . 27 = 5,4g$

c) $m_{H_2} = n . M = 0,3 . 2 = 0,6g$

$m_{dd\text{ sau pứ}} = 300 + 5,4 - 0,6 = 304,8g$
Theo PTHH:

$n_{Al_2(SO_4)_3} = \dfrac{1}{3}n_{H_2SO_4} = \dfrac{1}{3} . 0,3 = 0,1\text{ mol}$

$\Rightarrow m_{Al_2(SO_4)_3} = n . M = 0,1 . [2 . 27 + 3 . (32 + 4 . 16)] = 0,1 . 342 = 34,2g$

$\Rightarrow C\%_{Al_2(SO_4)_3} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{34,2}{304,8} . 100\% \approx 11,22\%$