a) PTHH: Fe + 2HCl \(\rightarrow\) FeCl2 + H2\(\uparrow\)
b) mHCl = \(\frac{20.100}{100}=20\left(g\right)\)
=> nHCl = \(\frac{20}{36,5}=\frac{40}{73}\left(mol\right)\)
Theo PT: n\(H_2\) = \(\frac{1}{2}\)nHCl = \(\frac{1}{2}\).\(\frac{40}{73}\) = \(\frac{20}{73}\)(mol)
=> V\(H_2\) = \(\frac{20}{73}\).22,4 \(\approx\)6,14 (l)
c) Theo PT: n\(FeCl_2\) = \(\frac{1}{2}\)nHCl = \(\frac{1}{2}\).\(\frac{40}{73}\) = \(\frac{20}{73}\)(mol)
=> m\(FeCl_2\) = \(\frac{20}{73}\).127 \(\approx\) 34,8 (g)