\(n_{HCl}=0,2\times2=0,4\left(mol\right)\),\(n_{Fe}=0,2\left(mol\right)\)
\(Fe+2HCl\rightarrow FeCl_2+H_2\uparrow\)
a , Theo PTHH ta có
\(n_{Fe}=n_{H_2}=0,2\left(mol\right)\Rightarrow V_{H_2}=\frac{0,2}{0,2}=1M\)
b , Ta có PTHH
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
Theo PTHH
\(n_{Al}=\frac{1}{3}n_{HCl}=\frac{2}{15}\left(mol\right)\Rightarrow m_{Al}=27\times\frac{2}{15}=3,6\left(g\right)\)