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