Đáp án:
\(\begin{array}{l} b,\ V_{H_2}=2,24\ lít.\\ c,\ C\%_{FeSO_4}=14,42\%\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l} a,\ PTHH:Fe+H_2SO_4\to FeSO_4+H_2↑\\ b,\ n_{Fe}=\dfrac{5,6}{56}=0,1\ mol\\ Theo\ pt:\ n_{H_2}=n_{Fe}=0,1\ mol.\\ \Rightarrow V_{H_2}=0,1\times 22,4=2,24\ lít.\\ c,\ Theo\ pt:\ n_{H_2SO_4}=n_{Fe}=0,1\ mol.\\ \Rightarrow m_{\text{dd H$_2$SO$_4$}}=\dfrac{0,1\times 98}{9,8\%}=100\ g.\\ Theo\ pt:\ n_{FeSO_4}=n_{Fe}=0,1\ mol.\\ ⇒m_{\text{dd spư}}=m_{Fe}+m_{\text{dd H$_2$SO$_4$}}-m_{H_2}\\ \Rightarrow m_{\text{dd spư}}=5,6+100-0,2=105,4\ g.\\ \Rightarrow C\%_{FeSO_4}=\dfrac{0,1\times 152}{105,4}\times 100\%=14,42\%\end{array}\)
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