Cho 8,654 gam hỗn hợp X gồm khí Cl2 và O2 tác dụng vừa đủ với 0,396 mol

* Hướng dẫn giải

Đáp án A

$\begin{array}{l}+ Sơ đồ phản ứng:\\ \underset{8,654 \left(g\right)}{\underbrace{\left\{\begin{array}{l}C{l}_2\\ O_2\end{array}\right\}}}+ \underset{0,396 mol}{\underbrace{\left\{\begin{array}{l}Mg, Al\\ Zn\end{array}\right\}}}\stackrel{t^0}{\to }\underset{23,246 \left(g\right)}{\underbrace{\left\{\begin{array}{l}oxit\\ muối\end{array}\right\}}}\underset{V ml}{\overset{HCl 1M}{\to }}\underset{T}{\underbrace{\left\{\begin{array}{l}M{g}^{2+}, A{l}^{3+}\\ Z{n}^{2+}, C{l}^{-}\end{array}\right\}}}\underset{0,572 mol}{\overset{Ba{\left(OH\right)}_2}{\to }}Mg(OH{)}_2\downarrow \\ + Phản ứng của T với Ba(OH{)}_2:\\ Mg+2O{H}^{-}\stackrel{}{\to }Mg(OH{)}_2\\ Z{n}^{2+}+4O{H}^{-}\stackrel{}{\to }Zn{O_2}^{2-}+2H_{2}O\\ A{l}^{3+}+4O{H}^{-}\stackrel{}{\to }Al{O_2}^{-}+2H_{2}O\\ + \left\{\begin{array}{l}{n}_{Mg}=x\\ n_{Zn}=y\\ n_{Al}=z\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_Y=x+y+z=0,396\\ m_Y=24x+65y+27z=23,246-8,654\\ 2x+4y+4z=n_{O{H}^{-}}=1,144\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,22\\ y=0,12\\ z=0,056\end{array}\right.\\ + \left\{\begin{array}{l}{m}_X=71n_{C{l}_2}+32n_{O_2}=8,654\\ BTE: 2n_{C{l}_2}+4n_{O_2}=0,22.2+0,12.2+0,056.3\end{array}\right.\Rightarrow \left\{\begin{array}{l}{n}_{C{l}_2}=0,034\\ n_{O_2}=0,195\end{array}\right.\\ + n_{H^{+}}=2n_{O^{2-}}=2.2n_{O_2}=0,78 mol\Rightarrow V_{HCl 1M}=0,78 lít=780 ml\end{array}$