Dẫn V lít (đktc) hỗn hợp X gồm axetilen và hiđro đi qua ống sứ đựng bột niken, nóng, thu được khí Y. Dẫn Y vào lượng dư AgNO3 trong dung dịch NH3 thu được 12 gam kết tủa

Trả lời

Khi cho Y qua lượng dư AgNO₃ trong dung dịch NH₃ thì có C₂H₂ phản ứng và tạo kết tủa

$$\begin{gathered} \text{HC}\equiv \text{CH}+2{\text{AgNO}}_3+2{\text{NH}}_3\to \text{AgC}\equiv \text{CAg}+2{\text{NH}}_4{\text{NO}}_3{\text{n}}_{{\text{C}}_2{\text{H}}_2}={\text{n}}_{{\text{C}}_2{\text{Ag}}_2}=\frac{12}{240}=0,05 \text{mol} \\ {\text{C}}_2{\text{H}}_4+{\text{Br}}_2\to {\text{C}}_2{\text{H}}_4{\text{Br}}_2 \\ {\text{n}}_{{\text{C}}_2{\text{H}}_4}={\text{n}}_{{\text{Br}}_2}=\frac{16}{160}=0,1 \text{mol} \end{gathered}$$

Khí Z gồm C₂H₆ (a mol) và H₂ dư (b mol)

$\begin{array}{l}{\text{C}}_2{\text{H}}_6+\frac{7}{2}{\text{O}}_2\stackrel{{\text{t}}^0}{\to }2{\text{CO}}_2+3{\text{H}}_2\text{O}\\ amol2amol3amol\\ 2{\text{H}}_2+{\text{O}}_2\stackrel{{\text{t}}^0}{\to }2{\text{H}}_2\text{O}\\ \text{b mol\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}b mol}\end{array}$           $\left\{\begin{array}{l}2a=\frac{2,24}{22,4}\\ 3a+b=\frac{4,5}{18}\end{array}\to \left\{\begin{array}{l}a=0,05\\ b=0,1\end{array}\right.\right.$

Trong hỗn hợp X ban đầu:

(Bảo toàn C): ${\text{n}}_{{\text{C}}_2{\text{H}}_2}={\text{n}}_{{\text{C}}_2{\text{H}}_2\left(\text{Y}\right)}+{\text{n}}_{{\text{C}}_2{\text{H}}_4}+{\text{n}}_{{\text{C}}_2{\text{H}}_6}=0,05+0,1+0,05=0,2 \text{mol}$

(Bảo toàn H): ${\text{n}}_{{\text{C}}_2{\text{H}}_2}+{\text{n}}_{{\text{H}}_2}={\text{n}}_{{\text{C}}_2{\text{H}}_2\left(\text{Y}\right)}+2\cdot {\text{n}}_{{\text{C}}_2{\text{H}}_4}+3{\text{n}}_{{\text{C}}_2{\text{H}}_6}+{\text{n}}_{{\text{H}}_2\left(\text{Y}\right)}\to {\text{n}}_{{\text{H}}_2}=0,3 \text{mol}$

$\to {\text{n}}_{\text{x}}={\text{n}}_{{\text{C}}_2{\text{H}}_2}+{\text{n}}_{{\text{H}}_2}=0,2+0,3=0,5 \text{mol}\to \text{V}=0,5\cdot 22,4=11,2$ (lít).