Hòa tan hoàn toàn 8 gam hỗn hợp X gồm

Chọn đáp án A

Giải thích:$\begin{array}{l}+\left\{\begin{array}{l}F\text{eS:\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}mol}\\ F{\text{eS}}_2:ymol\end{array}\right\}\stackrel{HN{O}_3}{\to }\left\{\begin{array}{l}F{\text{e}}^{3+};N{O_3}^{-}\\ S{O_4}^{2-};H^{+}\end{array}\right\}\stackrel{\mathrm{0,25}molBa{\left(OH\right)}_2}{\to }\left\{\begin{array}{l}B\text{aS}{O}_4\\ Fe{\left(OH\right)}_3\end{array}\right\}\stackrel{t^o}{\to }\left\{\begin{array}{l}B\text{aS}{O}_4:(x+2y)mol\\ F{\text{e}}_2{O}_3:\mathrm{0,5}(x+y)mol\end{array}\right\}\\ \Rightarrow \left\{\begin{array}{l}88\text{x}+120y=8\\ 233(x+2y)+\mathrm{160.0,5}(x+y)=\mathrm{32,03}\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=\mathrm{0,05}\\ y=\mathrm{0,03}\end{array}\right.\\ \Rightarrow n_{HN{O}_3}=2n_{Ba{\left(N{O}_3\right)}_2}+n_{NO}=2(\mathrm{0,25}-\mathrm{0,11})+\frac{\mathrm{0,05.9}+\mathrm{0,03.15}}{3}=\mathrm{0,58}mol\Rightarrow \boxed{V_{\text{dd\hspace{0.17em}\hspace{0.17em}}HN{O}_{3}2M}=290ml}\end{array}$