a,

$n_{Fe(OH)_2}=\dfrac{18}{90}=0,2 mol$

$8Fe(OH)_2+13H_2SO_4\to 4Fe_2(SO_4)_3+H_2S+20H_2O$

$\Rightarrow n_{H_2S}=0,025 mol$

$p=540mmHg=\dfrac{27}{38}(atm)$

$\Rightarrow V=\dfrac{0,025.0,082.(27+273)}{\dfrac{27}{38}}=0,865l$

b,

$n_{H_2SO_4}=\dfrac{0,2.13}{8}=0,325 mol$

$\Rightarrow m_{dd H_2SO_4}=0,325.98:80\%=39,8125g$

$\Rightarrow m_X=18+39,8125-0,025.34=56,9625g$

$n_{Fe_2(SO_4)_3}=0,2:2=0,1 mol$

$\to C\%_{Fe_2(SO_4)_3}=\dfrac{0,1.400.100}{56,9625}=70,22\%$

Đáp án:

a/ $V≈0,866\ lít$

 b/ $C\%=70,22\%$

Giải thích các bước giải:

 Khí sinh ra mùi trứng thối ⇒ là $H_2S$

$8Fe(OH)_2+13H_2SO_4\to 4Fe_2(SO_4)_3+H_2S↑+20H_2O$

$n_{Fe(OH)_2}=\dfrac{18}{90}=0,2\ mol$

a/ Theo PTHH 

$n_{H_2S}=\dfrac{1}{8}.n_{Fe(OH)_2}=0,025\ mol$

$540 mmHg = \dfrac{540}{760}atm =\dfrac{27}{38}atm$

$⇒V=\dfrac{n.R.T}{P}\\⇔V=\dfrac{0,025.0,082.(27+273)}{\dfrac{27}{38}}≈0,866\ lít$

b/ Theo PTHH: 

$n_{H_2SO_4}=\dfrac{13}{8}.n_{Fe(OH)_2}=0,325\ mol\\⇒m_{H_2SO_4}=0,325.98=31,85g\\⇒m_{dd\ H_2SO_4}=39,8125g$

$⇒m_X=18+39,8125-0,025.34=56,9625g$

$n_{Fe_2(SO_4)_3}=0,5.0,2=0,1\ mol\\⇒m_{Fe_2(SO_4)_3}=40g\\⇒C\%=\dfrac{40}{56,9625}.100\%=70,22\%$