\(Fe+2HCl\rightarrow FeCl_2+H_2\)
a)
\(n_{Fe}=\frac{11,2}{22,4}=0,2\left(mol\right)\)
\(n_{Fe}=n_{Fe}=0,2\left(mol\right)\)
\(\Rightarrow V_{H2}=0,2.2,4=4,48\left(l\right)\)
b)
\(n_{HCl}=2n_{Fe}=2.0,2=0,4\left(mol\right)\)
\(m_{HCl}=0,4.36,5=14,6\left(g\right)\)
c)
\(n_{FeCl2}=n_{Fe}=0,2\left(mol\right)\)
\(m_{FeCl2}=0,2.127=25,4\left(g\right)\)