Đáp án + Giải thích các bước giải:

 Câu 3:
Tóm tắt:
$m_{Mg} = 2,4g$
$m_{ddHCl} = 200g$
a) $C\%_{HCl} = ?\%$

b) $m_{ZnCl_2} = ?g$

c) $V_{H_2} = ?l$

Giải:
a) $n_{Mg} = \dfrac{m}{M} = \dfrac{2,4}{24} = 0,1(mol)$

PTHH:
$Mg  + 2HCl \longrightarrow MgCl_2 + H_2 \uparrow$
Theo PTHH:
$n_{HCl} = 2n_{Mg} = 2. 0,1 = 0,2(mol)$
$\Rightarrow m_{HCl} =n . M = 0,2 . (35,5 + 1) = 0,2 . 36,5 = 7,3(g)$

$\Rightarrow C\%_{HCl} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{7,3}{200} . 100\% = 3,65\%$

b) Theo PTHH:
$n_{MgCl_2} = n_{Mg} = 0,1(mol)$

$\Rightarrow m_{MgCl_2} =n . M = 0,1 . (24 + 35,5 . 2) = 0,1 . 95 = 9,5(g)$

c) Theo PTHH:
$n_{H_2} = n_{Mg} = 0,1(mol)$
$\Rightarrow V_{H_2} =n . 22,4 = 0,1 . 22,4 = 2,24l$

Câu 4:

Tóm tắt:

$m_{Zn} = 3,25g$

$V_{HCl} = 200ml = 0,2l$

a) $C_{MHCl} = ?M$

b) $m_{ZnCl_2} = ?g$

c) $V_{H_2} = ?l$

Giải:
a) $n_{Zn} = \dfrac{m}{M} = \dfrac{3,25}{65} = 0,05(mol)$

PTHH:
$Zn + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$

Theo PTHH:
$n_{HCl} = 2n_{Zn} = 2 . 0,05 = 0,1(mol)$
$\Rightarrow C_{MHCl} = \dfrac{n}{V} = \dfrac{0,1}{0,2} = 0,5M$

b) Theo PTHH:
$n_{ZnCl_2} = n_{Zn} = 0,05(mol)$

$\Rightarrow m_{ZnCl_2} =n . M = 0,05 . (65 + 2. 35,5) = 0,05 . 136 = 6,8(g)$
c) Theo PTHH:
$n_{H_2} =n_{Zn} = 0,05\text{ mol}$

$\Rightarrow V_{H_2} =n . 22,4 = 0,05 . 22,4 = 1,12l$