PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\)
Ta có: \(n_{H_2}=\dfrac{6,72}{22,4}=0,3\left(mol\right)=n_{Fe}=n_{FeSO_4}\)
\(\Rightarrow\left\{{}\begin{matrix}m_{Hg}=30-0,3\cdot56=13,2\left(g\right)\\m_{FeSO_4}=0,3\cdot152=45,6\left(g\right)\\m_{H_2}=0,3\cdot2=0,6\left(g\right)\\\end{matrix}\right.\)
\(\Rightarrow m_{dd}=m_{hh}+m_{ddH_2SO_4}-m_{H_2}-m_{Hg}=76,2\left(g\right)\)
\(\Rightarrow C\%_{FeSO_4}=\dfrac{45,6}{76,2}\cdot100\%\approx59,84\%\)
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