Rót từ từ dung dịch chứa a mol...

C

- Hướng dẫn giải

Đáp án C

$\begin{array}{l}+ TN1: n_{C{O}_2}=n_{H^{+}}-n_{C{O_3}^{2-}}\Rightarrow \frac{a+b}{7}=a-b\Rightarrow \left\{\begin{array}{l}6a-8b=0\\ a>b\end{array}\right.\\ X có \left\{\begin{array}{l}{n}_{N{a}^{+}}=a+2b\\ n_{C{l}^{-}}=a; n_{HC{O_3}^{-}}=2b\end{array}\right.\Rightarrow m_{chất \tan trong X}=23(a+2b)+35,5a+2b.61\\ + TN2: C{O}_2+ \left\{\begin{array}{l}N{a}_{2}C{O}_3\\ NaOH\end{array}\right.\iff \underset{a mol}{\underbrace{H_{2}C{O}_3}}+ \left\{\begin{array}{l}N{a}_{2}C{O}_3: a mol\\ NaOH: b mol\end{array}\right.\left\{\begin{array}{l}NaHC{O}_3\\ N{a}_{2}C{O}_3\end{array}\right.+\underset{b mol}{\underbrace{HOH}}\\ \Rightarrow m_{chất \tan trong Y}=62a+106a+40b-18b\\ \Rightarrow m_{chất \tan trong X và Y}=226,5a+190b=59,04\\ \Rightarrow \left\{\begin{array}{l}6a-8b=0\\ 226,5a+190b=59,04\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=0,16\\ b=0,12\end{array}\right.\\ \Rightarrow Y chứa \left\{\begin{array}{l}NaHC{O}_3: y\\ N{a}_{2}C{O}_3: x\end{array}\right.\Rightarrow \left\{\begin{array}{l}BTNT Na: 2x+y=0,44\\ BTNT C: x+y=0,32\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=0,12\\ y=0,2\end{array}\right.\\ \Rightarrow Y+BaC{l}_2 dư\stackrel{}{\to }0,12 mol BaC{O}_3\Rightarrow m_{BaC{O}_3}=23,64 gam\end{array}$