Đáp án + Giải thích các bước giải:

a) PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$
$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2 \uparrow$

b) $n_{H_2} = \dfrac{V}{22,4} = \dfrac{20,16}{22,4} = 0,9\text{ mol}$

Theo PTHH:
$n_{H_2} = n_{Zn} + \dfrac{3}{2}n_{Al} = 0,9\text{ mol}$

Mà $n_{Zn} = n_{Al}$

$\Rightarrow n_{H_2} = n_{Zn} + \dfrac{3}{2}n_{Zn} = \dfrac{5}{2}n_{Zn} = 0,9\text{ mol}$

$\Rightarrow n_{Zn} =0,9 : \dfrac{5}{2} = 0,36\text{ mol}$

$\Rightarrow n_{Al} = n_{Zn} = 0,36\text{ mol}$

$m_{Zn} = n . M = 0,36 . 65 = 23,4g$

$m_{Al} = n . M = 0,36 . 27 = 9,72g$

$\%_{Zn} = \dfrac{23,4}{23,4 + 9,72} \approx 70,65\%$

$\%_{Al} = 100 - 70,65\% = 29,35\%$

c) Theo PTHH:
$n_{HCl} = 2n_{Zn} + 3n_{Al} = 2n_{Zn} + 3n_{Zn} = 5n_{Zn} = 5 . 0,36 = 1,8\text{ mol}$

$\Rightarrow m_{HCl} = n . M = 1,8 . 36,5 = 65,7g$

$V_{H_2} = 20,16l$(đã cho trên đề bài)

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