a) TN3: Cho B td HCl

pt tạo khí:

Fe + HCl -------> FeCl₂ + H₂

0,05____________________\(_{\leftarrow}\)0,05

Đặt \(n_{FeO}=a;n_{Fe_2O_3}=b\left(mol\right)\)

BTNT.Fe \(\Rightarrow0,05+a+2b=2n_{Fe_2O_3\left(bd\right)}=0,2\left(1\right)\)

TN4: Cho D td NaOH

Do n_NaOH < n_Cl^- => NaOH dư

\(\Rightarrow n_{Fe\left(OH\right)_2}=a+0,05\left(mol\right);n_{Fe\left(OH\right)_3}=2b\left(mol\right)\\ \Rightarrow m_{\downarrow}=90\left(a+0,05\right)+214b=18,34\left(2\right)\)

(1) và (2) \(\Rightarrow\left\{{}\begin{matrix}a=0,13\\b=0,01\end{matrix}\right.\)

Từ đó suy ra 5 khối lượng các chất.

b) TN1: Cho CO td Fe₂O₃

Đặt \(n_{CO}=x;n_{CO_2}=y\left(mol\right)\)

BTNT.C \(\Rightarrow x+y=n_{CO\left(bd\right)}=0,3\left(3\right)\)

BTNT.O \(\Rightarrow n_{CO}+2n_{CO_2}+n_{FeO}+3n_{Fe_2O_3}=3n_{Fe_2O_3\left(bd\right)}+n_{CO\left(bd\right)}\)

\(\Rightarrow x+2y+0,13+0,03=0,3+0,3\left(4\right)\)

(3) và (4) => \(\left\{{}\begin{matrix}x=0,16\\y=0,14\end{matrix}\right.\)

\(d_{A/k^2}=\frac{0,16\cdot28+0,14\cdot44}{0,3\cdot29}=1,22\)

c) Cho A td Ba(OH)₂

\(n_{BaCO_3}=0,06\left(mol\right)< n_{CO_2}\)

=> Đã có pứ hòa tan ↓

Đặt \(n_{Ba\left(OH\right)_2}=c\left(mol\right)\)

Ba(OH)₂ + CO₂ --------> BaCO₃ + H₂O

c\(_{\rightarrow}\)___________c________c

BaCO₃ + CO₂ + H₂O --------> Ba(HCO₃)₂

0,14-c______\(_{\leftarrow}\)0,14-c

\(\Rightarrow n_{BaCO_3\left(con\right)}=c-\left(0,14-c\right)=0,06\\ \Rightarrow c=0,1\Rightarrow C_{M\left(Ba\left(OH\right)_2\right)}=1\left(M\right)\)