Đáp án + Giải thích các bước giải:

 $n_{Zn} = \dfrac{m}{M} = \dfrac{6,5}{65} = 0,1(mol)$

$m_{HCl} = \dfrac{C\% . m_{dd}}{100\%} = \dfrac{7,3\% . 150}{100\%} = 10,95(g)$

$n_{HCl} = \dfrac{m}{M} = \dfrac{10,95}{1 + 35,5} = \dfrac{10,95}{36,5} = 0,3(mol)$
PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$
Ta có:
$n_{Zn} < \dfrac{n_{HCl}}{2}\bigg(0,1 < \dfrac{0,3}{2}\bigg)$
$\Rightarrow HCl$ dư, tính theo $Zn$
Theo PTHH:
$n_{HCl(pứ)} = 2n_{Zn} = 2 . 0,1 = 0,2(mol)$
$\Rightarrow n_{HCl(dư)} = 0,3 - 0,2 = 0,1(mol)$
$m_{HCl(dư)} = n . M = 0,1 . 36,5 = 3,65(g)$

Theo PTHH:

$n_{H_2} = n_{ZnCl_2} = n_{Zn} = 0,1(mol)$
$\Rightarrow m_{H_2} = n . M = 0,1 . 2 = 0,2(g)$

$m_{ZnCl_2} = n . M = 0,1 . (65 + 2 .35,5) = 0,1 . 136 = 13,6(g)$

$m_{dd\text{sau pứ}} = m_{Zn} + m_{ddHCl} - m_{H_2} = 6,5 + 150 -  0,2 = 156,3(g)$

$\Rightarrow C\%_{ZnCl_2} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{13,6}{156,3} . 100\% \approx 8,7\%$

$C\%_{HCl(dư)}= \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{3,65}{156,3} . 100\% \approx 2,34\%$