Đáp án + Giải thích các bước giải:

 Bài 6:

a) $n_{H_2} = \dfrac{V}{22,4} =\dfrac{0,672}{22,4} = 0,03\text{ mol}$

PTHH:
$2Al + 6HCl \longrightarrow 2AlCl_3 + 3H_2 \uparrow$
Theo PTHH:
$n_{Al} = \dfrac{2}{3}n_{H_2} = \dfrac{2}{3} . 0,03 = 0,02\text{ mol}$

$\Rightarrow m_{Al} = n . M = 0,02 . 27 = 0,54g$

b) Theo PTHH:
$n_{HCl} = 2n_{H_2} = 2 . 0,03 = 0,06\text{ mol}$

$\Rightarrow m_{HCl} = n . M = 0,06 . (35,5 + 1) =0,06 . 36,5 = 2,19g$

$\Rightarrow m = m_{ddHCl} = \dfrac{m_{ct} . 100\%}{C\%} = \dfrac{2,19 .100\%}{7,3\%} = 30g$

c) $m_{H_2}=  n . M = 0,03 . 2 = 0,06g$
Theo PTHH:

$n_{AlCl_3} = n_{Al} = 0,02\text{ mol}$
$\Rightarrow m_{AlCl_3} = n . M = 0,02 . (27 + 35,5 . 3) = 0,02 . 133,5 = 2,67g$

$m_{ddAlCl_3} = m_{Al} + m_{ddHCl} - m_{H_2} = 0,54 + 30 - 0,06 = 30,48g$
$\Rightarrow C\%_{AlCl_3} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{2,67}{30,48} .100\% \approx 8,76\%$

Bài 7:

$200ml = 0,2l$
$n_{Zn}= \dfrac{m}{M} = \dfrac{13}{65} = 0,2\text{ mol}$
PTHH:
$Zn + H_2SO_4 \longrightarrow ZnSO_4 + H_2 \uparrow$

Theo PTHH:
$n_{ZnSO_4} = n_{Zn} = 0,2\text{ mol}$
$\Rightarrow C_{MZnSO_4} = \dfrac{n}{V} = \dfrac{0,2}{0,2} = 1M$

Bài 8:
$m_{HCl} = \dfrac{C\% . m_{dd}}{100\%} = \dfrac{0,73\% . 400}{100\%} = 2,92g$

$n_{HCl} = \dfrac{m}{M} = \dfrac{2,92}{35,5 + 1} = \dfrac{2,92}{36,5} = 0,08\text{ mol}$

PTHH:
$R + 2HCl \longrightarrow RCl_2 + H_2 \uparrow$
Theo PTHH:
$n_R = \dfrac{1}{2}n_{HCl} = \dfrac{1}{2} . 0,08 = 0,04\text{ mol}$
$\Rightarrow M_R = \dfrac{m}{n} = \dfrac{2,6}{0,04} = 65g/mol = M_{Zn}$
Vậy $R$ là $Zn$