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Câu hỏi: Hấp thụ hoàn toàn 3,36 lít CO2 (đktc) vào dung dịch chứa a mol NaOH và b mol Na2CO3, thu được dung dịch X. Chia X thành hai phần bằng nhau, cho từ từ phần một vào 120 ml dung dịch HCl 1M, thu được 2,016 lít CO2 (đktc). Cho phần hai phản ứng hết với dung dịch Ba(OH)2 dư, thu được 29,55 gam kết tủa. Tỉ lệ $a:b$ tương ứng là
A. $2:1$.
B. $2:5$.
C. $1:2$.
D. $2:3$.
A. $2:1$.
B. $2:5$.
C. $1:2$.
D. $2:3$.
Nhận xét:
$C{{O}_{2}}\xrightarrow{NaOH,N{{a}_{2}}C{{O}_{3}}}X\left[ \begin{aligned}
& NaO{{H}_{du}}, N{{a}_{2}}C{{O}_{3}}\xrightarrow{HCl}C{{O}_{2}}\to \dfrac{{{n}_{HCl}}}{{{n}_{C{{O}_{2}}}}}\ge 2 \\
& N{{a}_{2}}C{{O}_{3}}, NaHC{{O}_{3}}\xrightarrow{HCl}C{{O}_{2}}\to 2\ge \dfrac{{{n}_{HCl}}}{{{n}_{C{{O}_{2}}}}}\ge 1 \\
\end{aligned} \right.$
Theo đề bài: ${{n}_{C{{O}_{2}}}}=0,09, {{n}_{HCl}}=0,012\to \dfrac{{{n}_{HCl}}}{{{n}_{C{{O}_{2}}}}}=1,33$
Vậy dung dịch A chứa Na2CO3, NaHCO3
Khi cho $\dfrac{1}{2}X$ vào HCl thì:
$\left\{ \begin{aligned}
& {{n}_{HCO_{3}^{-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}+2{{n}_{CO_{3}^{2-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}={{n}_{{{H}^{+}}}}=0,12 \\
& {{n}_{HCO_{3}^{-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}+{{n}_{CO_{3}^{2-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}=0,09 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& {{n}_{HCO_{3}^{-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}=0,06mol \\
& {{n}_{CO_{3}^{2-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}=0,03mol \\
\end{aligned} \right.\to \dfrac{{{n}_{HCO_{3}^{2}}}}{{{n}_{CO_{3}^{2-}}}}=2$
Khi cho $\dfrac{1}{2}X$ vào Ba(OH)2 dư thì: ${{n}_{HCO_{3}^{-}}}+{{n}_{CO_{3}^{2-}}}={{n}_{BaC{{O}_{3}}}}=0,15\to \left\{ \begin{aligned}
& {{n}_{HCO_{3}^{-}}}=0,1mol \\
& {{n}_{CO_{3}^{2-}}}=0,05mol \\
\end{aligned} \right.$
$\to $ X chứa $CO_{3}^{2-}$ (0,1 mol), $HCO_{3}^{-}$ (0,2 mol), Na+ ( $a+2b$ mol).
$\xrightarrow{BT:C}0,15+b=0,3\Rightarrow b=0,15mol\xrightarrow{BTDT\left( Y \right)}a=0,1mol$
$\to a:b=2:3$.
$C{{O}_{2}}\xrightarrow{NaOH,N{{a}_{2}}C{{O}_{3}}}X\left[ \begin{aligned}
& NaO{{H}_{du}}, N{{a}_{2}}C{{O}_{3}}\xrightarrow{HCl}C{{O}_{2}}\to \dfrac{{{n}_{HCl}}}{{{n}_{C{{O}_{2}}}}}\ge 2 \\
& N{{a}_{2}}C{{O}_{3}}, NaHC{{O}_{3}}\xrightarrow{HCl}C{{O}_{2}}\to 2\ge \dfrac{{{n}_{HCl}}}{{{n}_{C{{O}_{2}}}}}\ge 1 \\
\end{aligned} \right.$
Theo đề bài: ${{n}_{C{{O}_{2}}}}=0,09, {{n}_{HCl}}=0,012\to \dfrac{{{n}_{HCl}}}{{{n}_{C{{O}_{2}}}}}=1,33$
Vậy dung dịch A chứa Na2CO3, NaHCO3
Khi cho $\dfrac{1}{2}X$ vào HCl thì:
$\left\{ \begin{aligned}
& {{n}_{HCO_{3}^{-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}+2{{n}_{CO_{3}^{2-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}={{n}_{{{H}^{+}}}}=0,12 \\
& {{n}_{HCO_{3}^{-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}+{{n}_{CO_{3}^{2-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}=0,09 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& {{n}_{HCO_{3}^{-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}=0,06mol \\
& {{n}_{CO_{3}^{2-}\text{pha }\!\!\hat{\mathrm{u}}\!\!\text{ n }\!\!\ddot{\mathrm{o}}\!\!\text{ }\!\!\grave{\mathrm{u}}\!\!\text{ ng}}}=0,03mol \\
\end{aligned} \right.\to \dfrac{{{n}_{HCO_{3}^{2}}}}{{{n}_{CO_{3}^{2-}}}}=2$
Khi cho $\dfrac{1}{2}X$ vào Ba(OH)2 dư thì: ${{n}_{HCO_{3}^{-}}}+{{n}_{CO_{3}^{2-}}}={{n}_{BaC{{O}_{3}}}}=0,15\to \left\{ \begin{aligned}
& {{n}_{HCO_{3}^{-}}}=0,1mol \\
& {{n}_{CO_{3}^{2-}}}=0,05mol \\
\end{aligned} \right.$
$\to $ X chứa $CO_{3}^{2-}$ (0,1 mol), $HCO_{3}^{-}$ (0,2 mol), Na+ ( $a+2b$ mol).
$\xrightarrow{BT:C}0,15+b=0,3\Rightarrow b=0,15mol\xrightarrow{BTDT\left( Y \right)}a=0,1mol$
$\to a:b=2:3$.
Đáp án D.