Hỗn hợp X gồm M và R2O trong đó M

Chọn đáp án B

Giải thích:

$\begin{array}{l}+n_{HCl}=\frac{\mathrm{58,4.12}\%}{\mathrm{36,5}}=\mathrm{0,192}mol.\\ Vìchuabietchat\tan gomnhunggìnentaxet2truonghop\\ •TH1:Yco\left\{\begin{array}{l}MC{l}_2:xmol\\ RCl:xmol\end{array}\right.\\ +BTNTCl:3x=\mathrm{0,192}\Rightarrow x=\mathrm{0,064}\Rightarrow \mathrm{0,064}M+\mathrm{0,064}R=\mathrm{8,496}\left(loai\right).\\ •TH1:Yco\left\{\begin{array}{l}MC{l}_2:xmol\\ RCl:xmol\\ HCl:xmol\end{array}\right.\\ +BTNTCl:4x=\mathrm{0,192}\Rightarrow x=\mathrm{0,048}\Rightarrow \mathrm{0,048}M+\mathrm{0,048}R=\mathrm{8,448}\\ \Rightarrow M=137\left(Ba\right);R=39\left(K\right)\Rightarrow m=\mathrm{8,332}gam.\end{array}$