Đáp án đúng: D
Fe3O4 và 360.
Cho khí A (CO2) hấp thụ vào Ba(OH)2 :
$\displaystyle \xrightarrow{\text{BTNT}\text{.C}}{{\text{n}}_{\text{C}{{\text{O}}_{\text{2}}}}}\text{=}{{\text{n}}_{\text{FeC}{{\text{O}}_{\text{3}}}}}\text{=}{{\text{n}}_{\text{BaC}{{\text{O}}_{\text{3}}}}}\text{=0,04(mol)}$
Ta có:
$\displaystyle {{\text{n}}_{\text{F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}\text{=0,14(mol)}\xrightarrow{\text{BTNT}\text{.Fe}}\sum{{{\text{n}}_{\text{Fe}}}}\text{=0,28(mol)}$
$\displaystyle \xrightarrow{\text{BTNT}\text{.Fe}}\text{n}_{\text{Fe}}^{\text{trong}\,\,\text{F}{{\text{e}}_{\text{x}}}{{\text{O}}_{\text{y}}}}\text{=0,28-0,04=0,24(mol)}$
$\displaystyle \xrightarrow{\text{BTKL}}{{\text{m}}_{\text{F}{{\text{e}}_{\text{x}}}{{\text{O}}_{\text{y}}}}}\text{=23,2-0,04}\text{.116=18,56(gam)}$
$\displaystyle \Rightarrow \text{n}_{\text{O}}^{\text{trong}\,\,\text{oxit}}\text{=}\frac{\text{18,56-0,24}\text{.56}}{\text{16}}\text{=0,32(mol)}$
Với FexOy ta có
$\displaystyle \frac{\text{x}}{\text{y}}\text{=}\frac{\text{0,24}}{\text{0,32}}\text{=}\frac{\text{3}}{\text{4}}\Rightarrow \text{F}{{\text{e}}_{\text{3}}}{{\text{O}}_{\text{4}}}\,\Rightarrow {{\text{n}}_{\text{FeO}\text{.F}{{\text{e}}_{\text{2}}}{{\text{O}}_{\text{3}}}}}\text{=0,08(mol)}\,\,\,\,\,\,\,\,\,\,\,\,$$\displaystyle \text{X+HCl}\Rightarrow \left\{ \begin{array}{l}\text{F}{{\text{e}}^{\text{2+}}}\text{:0,04+0,08=0,12(mol)}\,\,\,\\\text{F}{{\text{e}}^{\text{3+}}}\text{:0,08}\text{.2=0,16(mol)}\,\,\,\end{array} \right.\xrightarrow{\text{BTDT}}{{\text{n}}_{\text{C}{{\text{l}}^{\text{-}}}}}\text{=}{{\text{n}}_{\text{HCl}}}$
$\displaystyle \text{=0,12}\text{.2+0,16}\text{.3=0,72(mol)}\,\,\,$$\Rightarrow \text{V=}\frac{\text{720}}{\text{2}}\text{=360}\,\left( \text{ml} \right)$
