Đáp án:

b. \(\text V_{{\text H}_2}=1,344\ \text{l}\)

c. \( \text C\%_{H_2SO_4}=2,94\%\)

d. \(C\%_{FeSO_4}=4,487\%\)

Giải thích các bước giải:

a.  \[Fe+H_2SO_4\to FeSO_4+H_2\]

PTHH: 

b. 

\[m_{Fe}=3,36\ \text g\Rightarrow n_{Fe}=\dfrac{3,36}{56}=0,06\ \text{mol}\\n_{H_2}=n_{Fe}=0,06\ \text{mol}\Rightarrow \text V_{H_2}=0,06\times 22,4=1,344\ \text{l}\]

c. 

\[n_{H_2SO_4}= n_{Fe}=0,06\ \text{mol}\\\Rightarrow \text C\%_{H_2SO_4}=\dfrac{0,06\times 98}{200}\times 100\%=2,94\%\]

d. 

\[\text{BTKL}: m_{Fe}+m_{\text{dung dịch} \ H_2SO_4}=m_{\text{dung dịch sau phản ứng}}+m_{H_2}\\\Rightarrow m_{\text{dung dịch sau phản ứng}}=3,36+200-0,06\times 2=203,24\ \text{gam}\]

\[\Rightarrow C\%_{FeSO_4}=\dfrac{0,06\times 152}{203,24}\times 100\%=4,487\%\]

$n_{Fe}=3,36/56=0,06mol$

$a/Fe+H_2SO_4\to FeSO_4+H_2↑$

$\text{b/Theo pt :}$

$n_{H_2}=n_{Fe}=0,06mol$

$⇒V_{H_2}=0,06.22,4=1,344l$

$\text{c/Theo pt :}$

$n_{H_2SO_4}=n_{Fe}=0,06mol$

$⇒m_{H_2SO_4}=0,06.98=5,88g$

$⇒C\%_{H_2SO_4}=\dfrac{5,88}{200}.100\%=2,94\%$

$\text{d/Theo pt :}$

$n_{FeSO_4}=n_{Fe}=0,06mol$

$⇒m_{FeSO_4}=0,06.152=9,12g$

$m_{dd spu}=3,36+200-0,06.2=203,24g$

$⇒C\%_{FeSO_4}=\dfrac{9,12}{203,24\%}=4,48\%$