a) \(\begin{array}{l}{n_{Fe}} = \dfrac{{22,4}}{{56}} = 0,4(mol);{n_{{H_2}S{O_4}}} = \dfrac{{24,5}}{{98}} = 0,25(mol)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Fe + {H_2}S{O_4} \to FeS{O_4} + {H_2}\\PT(mol)\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\DB(mol)0,4\,\,\,\,\,\,\,\,\,\,0,25\\NX:0,25 < 0,4\end{array}\)
=> H2SO4 hết, Fe dư
\({n_{{H_2}}} = {n_{{H_2}S{O_4}}} = 0,25(mol) = > {V_{{H_2}}} = 0,25 \times 22,4 = 5,6(lit)\)
b) mFe dư = (0,4 - 0,25) x 56=8,4(g)
