a)\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
b)
\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PTHH:
\(n_{FeSO4}=n_{H2SO4}=n_{H2}=0,2\left(mol\right)\)
Ta có:
\(m_{FeSO4}=0,2.152=30,4\left(g\right)\)
\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)