a) Fe + 2HCl --> FeCl₂ + H₂

FeO + 2HCl --> FeCl₂ + H₂O

b) \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: Fe + 2HCl --> FeCl₂ + H₂

______0,5<-1<------0,5<---0,5

=> m_Fe = 0,5.56 = 28 (g)

=> \(\left\{{}\begin{matrix}\%Fe=\dfrac{28}{100}.100\%=28\%\\\%FeO=100\%-28\%=72\%\end{matrix}\right.\)

c) \(n_{FeO}=\dfrac{72}{72}=1\left(mol\right)\)

PTHH: FeO + 2HCl --> FeCl₂ + H₂O

______1---->2

=> m_HCl = (1+2).36,5 = 109,5 (g)

=> \(m_{ddHCl}=\dfrac{109,5.100}{30}=365\left(g\right)\)

=> \(V_{ddHCl}=\dfrac{365}{1,15}=317,39\left(ml\right)\)

a)

$Fe + 2HCl \to FeCl_2 + H_2$
$FeO +2 HCl \to FeCl_2 + H_2O$

b)

Theo PTHH : 

$n_{Fe} = n_{H_2} = \dfrac{4,48}{22,4} = 0,2(mol)$

$\%m_{Fe} = \dfrac{0,2.56}{20}.100\% = 56\%$

$\%m_{FeO} = 100\% - 56\% = 44\%$

c) $n_{FeO} = \dfrac{11}{90}(mol)$
$n_{HCl} = 2n_{Fe} + 2n_{FeO} = \dfrac{29}{45}(mol)$

$m_{dd\ HCl} = \dfrac{ \dfrac{29}{45}.36,5}{7,3\%} = 322,22(gam)$

a.\(Fe+2HCl->FeCl_2+H_2\)

b.\(FeO+2HCl->FeCl_2+H_2O\)

\(n_{H_2}=0,2\left(mol\right)\)

\(\%mFe=\dfrac{0,2.56}{20}.100\)

\(\%mFe=56\%\)

\(=>\%mFeO=100-56=44\%\)

\(n_{Fe}=\dfrac{2.8}{56}=0.05\left(mol\right)\)

\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

a) Chất tan : FeSO₄

Chất khí : H₂

\(m_{FeSO_4}=0.05\cdot152=7.6\left(g\right)\)

\(V_{H_2}=0.05\cdot22.4=1.12\left(l\right)\)

Fe + H2SO4 -> FeSO4 + H2

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A là: FeSO4

B là: H2

nFe = 2,8/56 = 0,05 mol => nH2SO4 = 0,05 mol

mFeSO4 = 0,05 x (56+32+64) = 7,6 gam 

VH2 = 0,05 x 22,4 = 1,12 lit