Dung dịch X gồm NaOH x mol/l và B...

B

- Hướng dẫn giải

Phương pháp giải:

+ Bảo toàn nguyên tố

+\({n_{O{H^ - }}} = 2{n_{C{O_3}^{2 - }}} + {n_{HC{O_3}^ - }}\)

Giải chi tiết:

Do dd M và dd N đều tác dụng với KHSO₄ sinh ra kết tủa nên dd M và dd N đều chứa Ba(HCO₃)₂

\(\begin{gathered}
TN1:200ml\,X\left\{ \begin{gathered}
NaOH\,xM \hfill \\
Ba{(OH)_2}\,yM \hfill \\
\end{gathered} \right.\xrightarrow{{0,04\,mol\,C{O_2}}}1,97g\,ket\,tua + {\text{dd}}\,M \hfill \\
TN2:200ml\,Y\left\{ \begin{gathered}
NaOH\,yM \hfill \\
Ba{(OH)_2}\,xM \hfill \\
\end{gathered} \right.\xrightarrow{{0,0325mol\,C{O_2}}}1,4775g\,ket\,tua + {\text{dd}}\,{\text{N}} \hfill \\
{\text{TN1:}}\, \hfill \\
{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}}}{\text{ = }}{{\text{n}}_{{\text{BaC}}{{\text{O}}_{\text{3}}}}}{\text{ = 0}}{\text{,01mol}} \hfill \\
{{\text{n}}_{{\text{HC}}{{\text{O}}_{\text{3}}}^ - }}{\text{ = }}{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}{\text{ - }}{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}}}{\text{ = 0}}{\text{,04 - 0}}{\text{,01 = 0}}{\text{,03mol [}}BTNT\,C{\text{]}} \hfill \\
\Rightarrow {n_{O{H^ - }}} = 2{n_{C{O_3}^{2 - }}} + {n_{HC{O_3}^ - }} \Leftrightarrow 0,2x + 0,4y = 0,05(1) \hfill \\
TN2: \hfill \\
{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}}}{\text{ = }}{{\text{n}}_{{\text{BaC}}{{\text{O}}_{\text{3}}}}}{\text{ = 0}}{\text{,0075mol}} \hfill \\
{{\text{n}}_{{\text{HC}}{{\text{O}}_{\text{3}}}^ - }}{\text{ = }}{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{2}}}}}{\text{ - }}{{\text{n}}_{{\text{C}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}}}{\text{ = 0}}{\text{,0325 - 0}}{\text{,0075 = 0}}{\text{,025mol [}}BTNT\,C{\text{]}} \hfill \\
\Rightarrow {n_{O{H^ - }}} = 2{n_{C{O_3}^{2 - }}} + {n_{HC{O_3}^ - }} \Leftrightarrow 0,2y + 0,4x = 0,04(2) \hfill \\
(1)\& (2) \Rightarrow \left\{ \begin{gathered}
x = 0,05M \hfill \\
y = 0,1M \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} \)

Đáp án B