a)
\(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)
PTHH: Fe + 2HCl --> FeCl2 + H2
0,15-->0,3----->0,15-->0,15
=> \(V_{H_2}=0,15.22,4=3,36\left(l\right)\)
b) \(m_{FeCl_2}=0,15.127=19,05\left(g\right)\)
c) \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
=> \(C\%=\dfrac{10,95}{200}.100\%=5,475\%\)