`a)PTHH:`
`Fe+2HCl->FeCl_2+H_2`
`0,1` `0,1` `(mol)`
`FeO+2HCl->FeCl_2+H_2 O`
`b)n_[H_2]=[2,24]/[22,4]=0,1(mol)`
`=>m_[Fe]=0,1.56=5,6(g)`
`=>m_[FeO]=10-5,6=4,4(g)`
\(a,n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
PTHH: Fe + 2HCl ---> FeCl2 + H2
0,1<-----------------------0,1
FeO + 2HCl ---> FeCl2 + H2O
b, \(\rightarrow\left\{{}\begin{matrix}m_{Fe}=0,1.56=5,6\left(g\right)\\m_{FeO}=10-5,6=4,4\left(g\right)\end{matrix}\right.\)