a) PTHH: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b) Ta có: \(\left\{{}\begin{matrix}n_{Fe}=\frac{3,5}{56}=0,0625\left(mol\right)\\n_{H_2SO_4}=\frac{100\cdot9,8\%}{98}=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow\) Fe phản ứng hết, H2SO4 còn dư
\(\Rightarrow n_{H_2}=0,0625mol\) \(\Rightarrow m_{H_2}=0,0625\cdot2=0,125\left(g\right)\)
\(\Rightarrow m_{dd}=m_{Fe}+m_{ddH_2SO_4}-m_{H_2}=103,375\left(g\right)\)
c) Theo PTHH: \(n_{FeSO_4}=n_{Fe}=0,0625mol\)
\(\Rightarrow m_{FeSO_4}=0,0625\cdot152=9,5\left(g\right)\)
d) Ta có: \(n_{H_2SO_4\left(dư\right)}=0,0375mol\)
\(\Rightarrow m_{H_2SO_4\left(dư\right)}=3,675\left(g\right)\) \(\Rightarrow C\%_{H_2SO_4\left(dư\right)}=\frac{3,675}{103,375}\cdot100\approx3,56\%\)