\(0,06\ mol \ X: \left\{\begin{matrix} CH_4 & \\ C_2H_2& \\ C_2H_4& \end{matrix}\right. + \overset{0,08 \ mol\ Br_2}{\longrightarrow} \left\{\be

Question

\(0,06\ mol \ X: \left\{\begin{matrix}
CH_4 & \\
C_2H_2& \\
C_2H_4&
\end{matrix}\right. + \overset{0,08 \ mol\ Br_2}{\longrightarrow} \left\{\begin{matrix}
C_2H_2Br_2 & \\
C_2H_2Br_4&
\end{matrix}\right.\\ 18\ gam\ X: \left\{\begin{matrix}
CH_4 & \\
C_2H_2& \\
C_2H_4&
\end{matrix}\right. \overset{+O_2}{\rightarrow} \left\{\begin{matrix}
CO_2 \overset{+Ca(OH)_2}{\longrightarrow} CaCO_3+H_2O& \\
H_2O &
\end{matrix}\right.\)
Khối lượng dung dịch giảm 54,48 gam so với ban đầu.
Tính \(d_{X/H_2}\)

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Vivian 1 năm 2021-10-02T20:20:23+00:00 1 Answers 12 views 0

Answers ( )

    0
    2021-10-02T20:21:41+00:00

    Gọi a, b, c là mol metan, axetilen, etilen trong 0,06 mol X.

    $\Rightarrow a+b+c=0,06$     (1) 

    $n_{Br_2}=0,08 mol$ 

    $\Rightarrow 2b+c=0,08$      (2) 

    Trong 18g X có ka, kb, kc mol mỗi chất.

    $\Rightarrow k(16a+26b+28c)=18$  (*)

    $CH_4+ 2O_2\to CO_2+2H_2O$

    $C_2H_2+\frac{5}{2}O_2\to 2CO_2+H_2O$

    $C_2H_4+3O_2\to 2CO_2+2H_2O$ 

    Theo PTHH, $n_{CO_2}= k(a+2b+2c)$; $n_{H_2O}= k(2a+b+2c)$

    $n_{CaCO_3}= n_{CO_2}=k(a+2b+2c)$ 

    $\Delta m= m_{CaCO_3}-m_{CO_2}-m_{H_2O}$ 

    $\Rightarrow k.[100(a+2b+2c)-44(a+2b+2c)-18(2a+b+2c)]=54,48$

    $\Leftrightarrow k(20a+94b+76c)=54,48$   (**)

    (**) $\Rightarrow k=\frac{18}{16a+26b+28c}=\frac{54,48}{20a+94b+76c}$ 

    $\Leftrightarrow 54,48(16a+26b+28c)= 18(20a+94b+76c)$

    $\Leftrightarrow 511,68a-275,52b+157,44c=0$   (3) 

    (1)(2)(3) $\Rightarrow a=0,01; b=0,03; c=0,02$ 

    $\overline{M}_X=\frac{0,01.16+0,03.26+0,02.28}{0,01+0,03+0,02}=25$

    $\Rightarrow d_{X/H_2}=12,5$

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