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trong 8,6 g X
Đặt $n_{CH4}=x, n_{C2H4}=y, n_{C2H2}=z$
$C2H4+Br2\rightarrow C2H4Br2$
y y
$C2H2+2Br2\rightarrow C2H2Br4$
z 2z
$\Rightarrow \begin{cases}16x+28y+26z=8,6 (1) \\ y+2z=\frac{48}{160} \end{cases}(2)$
trong 0,6 mol hỗn hợp X
$CH\equiv CH+2AgNO3+2NH3\rightarrow C2Ag2+NH4NO3$
$n_{C2Ag2}=0,15=n_{C2H2}=\frac{1}{4}n_{X}=\frac{1}{4}(x+y+z)\Rightarrow \frac{1}{4}+\frac{1}{4}+\frac{-3}{4}=0(3)$
(1)(2)(3)$\Rightarrow x=0,2, y=o,1, z=0,1$
%$V_{CH4} =\frac{0,2}{0,4}\times 100=50$%
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Trả lời 07-02-13 08:36 PM
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