Đáp án + Giải thích các bước giải:

Câu 29: A

 $m_{ddNaCl} = 225 + 25 = 250g$

$\Rightarrow C\%_{NaCl} = \dfrac{m_{ct}}{m_{dd}}  . 100\% = \dfrac{25}{250} . 100\% = 10\%$

Câu 36: Không đáp án nào đúng

$n_{Na} = \dfrac{m}{M} = \dfrac{2,3}{23} = 0,1\text{ mol}$

PTHH:
$2Na + 2H_2O \longrightarrow 2NaOH + H_2 \uparrow$
Theo PTHH:
$n_{H_2} = \dfrac{1}{2}n_{Na} = \dfrac{1}{2} . 0,1 = 0,05\text{ mol}$
$\Rightarrow V_{H_2} = n . 22,4 = 0,05 . 22,4 = 1,12l$

Câu 37: A

$m_{Na_2O} = n . M = 0,1 . (2 . 23 + 16) = 0,1 . 62 = 6,2g$

PTHH:
$Na_2O + H_2O \longrightarrow 2NaOH$

Theo PTHH:
$n_{NaOH} = 2n_{Na_2O} = 2 . 0,1 = 0,2\text{ mol}$

$\Rightarrow m_{NaOH} = n . M = 0,2 . (23 +16 + 1) = 0,2 . 40 = 8g$

$m_{ddNaOH} = m_{Na_2O} + m_{H_2O} = 500 + 6,2 = 506,2g$

$\Rightarrow C\%_{ddNaOH} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{8}{506,2} . 100\% \approx 1,58\%$

Câu 38: B
$n_{Fe} = \dfrac{m}{M} = \dfrac{11,2}{56} = 0,2\text{ mol}$

PTHH:
$Fe + 2HCl \longrightarrow FeCl_2 + H_2 \uparrow$

Theo PTHH:
$n_{HCl} = 2n_{Fe} = 2 . 0,2 = 0,4\text{ mol}$

$\Rightarrow m_{HCl} = n . M = 0,4 . (35,5 + 1) = 0,4 . 36,5 = 14,6g$

$\Rightarrow C\%_{HCl} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{14,6}{400} . 100\% = 3,65\%$

Câu 39: B
$n_{Zn} = \dfrac{m}{M} = \dfrac{13}{65} = 0,2\text{ mol}$

PTHH:
$Zn  + 2HCl \longrightarrow ZnCl_2 + H_2 \uparrow$
Theo PTHH:
$n_{H_2} = n_{Zn} = 0,2\text{ mol}$

$\Rightarrow V_{H_2} =n . 22,4 = 0,2 . 22,4 = 4,48l$

Câu 40: D
$n_{Fe}  = \dfrac{m}{M} = \dfrac{11,2}{56} = 0,2\text{ mol}$
PTHH:
$Fe + 2HCl \longrightarrow FeCl_2 + H_2 \uparrow$
Theo PTHH:
$n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2\text{ mol}$
$\Rightarrow m_{FeCl_2} =n . M = 0,2 . (56 + 2 . 35,5) = 0,2 . 127 = 25,4g$

$m_{H_2} =n . M = 0,2 . 2 = 0,4g$

$m_{ddFeCl_2} = m_{Fe} + m_{ddHCl} - m_{H_2}=  11,2 + 200 - 0,4 = 210,8g$

$\Rightarrow C\%_{FeCl_2} = \dfrac{m_{ct}}{m_{dd}} . 100\% = \dfrac{25,4}{210,8} . 100\% \approx 12,05\%$