a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)
Ta có: \(n_{H_2}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\)
b, \(n_{FeCl_2}=n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{FeCl_2}=0,4.127=50,8\left(g\right)\)
c, \(n_{HCl}=2n_{H_2}=0,8\left(mol\right)\)
\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)