a: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
\(n_{Fe}=\dfrac{5.6}{56}=0.1\left(mol\right)=n_{FeCl_2}\)
\(\Leftrightarrow n_{HCl}=2\cdot0.1=0.2\left(mol\right)\)
\(m_{HCl}=0.2\cdot36.5=7.3\left(g\right)\)
b: \(n_{H_2}=0.1\left(mol\right)\)
nên \(V_{H_2}=0.1\cdot22.4=2.24\left(lít\right)\)