a) n_{Fe} = \frac{12,6}{56} = 0,225mol
PTHH: Fe+2HCl \rightarrow FeCl_{2}+ H_{2}
n_{H_{2}}= \frac{0,225.1}{1}=0,225mol
V_{H_{2}}=0,025.22,4 = 5,04l
b) PTHH: Fe_{O}+H^{2} \rightarrow Fe + H_{2}O
n_{F_{e}} = \frac{0,225.1}{1}=0,225 mol
m_{F_{e}}=0,225.56=12,6g