Dung dịch X là dung dịch gồm Ba(OH)2 a mol/lít và NaOH b mol/lít. Hấp thụ

* Hướng dẫn giải

Theo đề ra ta có:

 $\begin{array}{l}{n}_{Ba{\left(OH\right)}_2}=0,2a\\ n_{NaOH}=0,2b\end{array}$;

Thí nghiệm 1: $\left\{\begin{array}{l}{\text{n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{5,6}}{\text{22,4}}\text{=0,25(mol)}\\ {\text{n}}_{\downarrow }{\text{=\hspace{0.17em}\hspace{0.17em}n}}_{{\text{BaCO}}_{\text{3}}}\text{=\hspace{0.17em}\hspace{0.17em}}\frac{\text{29,55}}{\text{197}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,15(mol)}\end{array}\right.$

Thí nghiệm 2:$\left\{\begin{array}{l}{\text{n}}_{{\text{CO}}_{\text{2}}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{10,08}{\text{22,4}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,45(mol)}\\ {\text{n}}_{\downarrow }{\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}n}}_{{\text{BaCO}}_{\text{3}}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{\text{9,85}}{\text{197}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0,05(mol)}\end{array}\right.$

Các phương trình hóa học xảy ra

Ba(OH)₂ + CO₂ $\underset{}{\to }$ BaCO₃↓ + H₂O

2NaOH + CO₂ $\underset{}{\to }$ Na₂CO₃ + H₂O

Na₂CO₃ + CO₂ + H₂O $\underset{}{\to }$2NaHCO₃

BaCO₃↓ + CO₂ + H₂O $\underset{}{\to }$ Ba(HCO₃)₂

Tại thí nghiệm 1: Kết tủa chưa bị tan

Nên ${\text{n}}_{\downarrow }{\text{=n}}_{{\text{Ba(OH)}}_{\text{2}}}\text{=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0,15(mol)}\Rightarrow \text{0,2a\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,15\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\Rightarrow \text{a\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,75 }$

Tại thí nghiệm 2: Kết tủa tan một phần nên

$\begin{array}{l}{\text{n}}_{\downarrow }{\text{=\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}2n}}_{{\text{Ba(OH)}}_{\text{2}}}{\text{\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}n}}_{\text{NaOH}}{\text{\hspace{0.17em}\hspace{0.17em}-\hspace{0.17em}\hspace{0.17em}n}}_{{\text{CO}}_{\text{2}}}\\ \text{⇒0,05\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}2.0,2a\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}\hspace{0.17em}0,2b\hspace{0.17em}\hspace{0.17em}-\hspace{0.17em}\hspace{0.17em}0,45}\end{array}$

Thay a = 0,75 ta được b = 1

Vậy a = 0,75; b = 1