\(\begin{array}{l}{n_{{H_2}S{O_4}}} = \dfrac{{49}}{{98}} = 0,5(mol)\\a)PTHH:\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Zn + {H_2}S{O_4} \to ZnS{O_4} + {H_2}\\PT(mol)\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\DB(mol)\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,0,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\end{array}\)
b) Theo PTHH trên axit H2SO4 còn dư, kim loại Zn hết sau phản ứng.
c) Tính thể tích khí hiđro thu được theo số mol kim loại kẽm:
\(\begin{array}{l}{n_{Zn}} = {n_{{H_2}}} = 0,2(mol)\\{V_{{H_2}}} = 0,2 \times 22,4 = 4,48(lit)\end{array}\)
