a)
Gọi : \(n_{Mg} = a(mol) ; n_{Zn} = b(mol) \Rightarrow 24a + 65b = 1,54(1)\\\)
\(Mg + 2HCl \to MgCl_2 + H_2\\
Zn + 2HCl \to ZnCl_2 + H_2\\
n_{H_2} = a + b = \dfrac{6,72}{22,4.} = 0,03(mol)(2)\)
Từ (1)(2) suy ra : a = 0,01 ; b = 0,02
\(\%m_{Mg} = \dfrac{0,01.24}{1,54}.100\% = 15,58\%\\
\%m_{Zn} = 100\% -15,88\%=84,12\%\)
b)
\(n_{HCl\ pư} = 2n_{H_2} = 0,06(mol) \Rightarrow n_{HCl\ dư} = \dfrac{0,06}{80\%}.20\% = 0,015(mol)\\
m_{dd\ sau\ pư} = 1,54 + 100 - 0,03.2 = 101,48(gam)\\
C\%_{MgCl_2} = \dfrac{0,01.95}{101,48}.100\% = 0,94\%\\
C\%_{ZnCl_2} = \dfrac{0,02.136}{101,48}.100\% = 2,68\%\\
C\%_{HCl} = \dfrac{0,015.36,5}{101,48}.100\% = 0,54\%\)