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.
\[\text{BTKL}: m_{Fe}+m_{\text{dung dịch} \ H_2SO_4}=m_{\text{dung dịch sau phản ứng}}+m_{H_2}\\\Rightarrow m_{\text{dung dịch sau phản ứng}}=3,36+200-0,06\times 2=203,24\ \text{gam}\]
\[\Rightarrow C\%_{FeSO_4}=\dfrac{0,06\times 152}{203,24}\times 100\%=4,487\%\]