Hỗn hợp X gồm (metan, etilen, axetilen và H2) trong đó số mol metan bằng số

* Hướng dẫn giải

a) ${\text{n}}_{{\text{Br}}_{\text{2}}}\text{= }\frac{\text{120}}{\text{160}}\text{ = 0,75(mol)}$

${\text{n}}_{hhX}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{\text{11,76}}{22,4}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0,525(mol)}$

${\text{n}}_{O_2}\text{=\hspace{0.17em}\hspace{0.17em}}\frac{\text{19,32}}{22,4}\text{\hspace{0.17em}\hspace{0.17em}= \hspace{0.17em}0,8625(mol)}$

Gọi số mol CH₄, C₂H₄ là x, số mol C₂H₂ là y, số mol H₂ là z (mol) trong 11,76 lít hỗn hợp X, khi đó trong 15,3 gam, số mol của CH₄, C₂H₄ là kx, số mol C₂H₂ là ky, số mol H₂ là kz (mol) (x, y, z > 0)

$\begin{array}{l}{C}_2{H}_4+B{r}_2\to C_2{H}_{4}B{r}_2\\ kx kx\end{array}$

$\begin{array}{l}{C}_2{H}_2+2B{r}_2\to C_2{H}_{2}B{r}_4\\ ky 2ky\end{array}$

$\begin{array}{l}C{H}_4+2O_2\stackrel{t°}{\to }C{O}_2+2H_{2}O\\ x 2x\end{array}$

$\begin{array}{l}{C}_2{H}_4+3O_2\stackrel{t°}{\to }2C{O}_2+2H_{2}O\\ x 3x\end{array}$

$\begin{array}{l}{C}_2{H}_2+O_2\stackrel{t°}{\to }2C{O}_2+H_{2}O\\ y y\end{array}$

$\begin{array}{l}2H_2+O_2\to 2H_{2}O\\ z \to \frac{z}{2} \end{array}$

Theo bài ra ta có:

$\begin{array}{l}\left\{\begin{array}{l}kx+2ky=0,75\\ k(44x+26y+2z)=15,3\\ 2x+y+z=0,525\\ 5x+\frac{5}{2}y+\frac{z}{2}=0,8625\end{array}\right.\\ hay\left\{\begin{array}{l}23,6x-14,8y+2z=0\\ 2x+y+z=0,525\\ 10x+5y+z=1,725\end{array}\right.\iff \left\{\begin{array}{l}x=0,075\\ y=0,15\\ z=0,225\end{array}\right.\end{array}$

${\text{\%V}}_{{\text{CH}}_{\text{4}}}{\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}\%V}}_{{\text{C}}_{\text{2}}{\text{H}}_{\text{4}}}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{\text{0,075}}{\text{0,525}}\text{.100\%\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em} 14,3\%}$

${\text{\%V}}_{{\text{C}}_{\text{2}}{\text{H}}_6}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{\text{0,15}}{\text{0,525}}\text{.100\%\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}28,6\%}$

${\text{\%V}}_{{\text{H}}_2}\text{\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}}\frac{\text{0,225}}{\text{0,525}}\text{.100\%\hspace{0.17em}\hspace{0.17em}=\hspace{0.17em} 42,8\%}$