a)\( Fe+2HCl\rightarrow FeCl_2+H_2\\Zn+2HCl\rightarrow ZmCl_2+H_2\\ n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Gọi a là số \(mol\) \(Fe\), b là số \(mol\) \(Zn\)
\(56a+65b=29,8\\ a+b=0,5\\ \Rightarrow a=0,3;b=0,2\\ \%m_{Fe}=\dfrac{0,3.56}{29,8.100\%}=56,38\%\\ \%m_{Zn}=00-56,8=43.62\%\)
b)\(n_{HCl}=0,5\times2=1\left(mol\right)\\ C_{M_{HCl}}=\dfrac{1}{0,6}=\dfrac{5}{3}M\)
c)Chịuuuu