\[\begin{array}{l}a){m_{Fe}} = \dfrac{{60,5 \times 46,289}}{{100}} \approx 28(g) = > {n_{Fe}} = \dfrac{{28}}{{56}} = 0,5(mol)\\{m_{Zn}} = 60,5 - 28 = 32,5(g) = > {n_{Zn}} = \dfrac{{32,5}}{{65}} = 0,5(mol)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Fe + 2HCl \to FeC{l_2} + {H_2}\\PT(mol)\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\DB(mol)\,\,\,0,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0,5 \to 0,5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Zn + 2HCl \to ZnC{l_2} + {H_2}\\PT(mol)\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\\DB(mol)\,\,\,0,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to 0,5 \to 0,5\\b)\sum {{n_{{H_2}}} = 0,5 + 0,5 = 1(mol) = > {V_{{H_2}}} = 1 \times 22,4 = 22,4(lit)} \\c)\,\,{m_{FeC{l_2}}} = 0,5 \times 127 = 63,5(g)\\\,{m_{ZnC{l_2}}} = 0,5 \times 136 = 68(g)\end{array}\]
Đáp số : \({V_{{H_2}}} = 22,4(l);{m_{FeC{l_2}}} = 63,5(g);{m_{ZnC{l_2}}} = 68(g).\)
