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$n_{anken}=\frac{10,08-8,4}{22,4}=0,075 mol\Rightarrow M_{anken}=\frac{3,15}{0,075}=42\Rightarrow anken:C_3H_6$
$C:\begin{cases}n_{H_2}=x mol \\n_{C_3H_8}=\frac{13,44-10,08}{22,4}=0,15 mol\\n_{C_nH_{2n+2}}=y mol \end{cases}$
$\Rightarrow \begin{cases}\frac{2x+44.0,15+(14n+2)y}{x+0,15+y}=2.17,8 (1)\\x+0,15+y=\frac{8,4}{22,4} (2)\end{cases}$
$(2)\Rightarrow x+y=0,225$
$(1)\Rightarrow ny=0,45\Rightarrow n\geqslant \frac{0,45}{0,225}=2$
$*)n=2\Rightarrow \begin{cases}x=0 \\ y=0,225 \end{cases}\Rightarrow A:\begin{cases}n_{H_2}=0,15 mol \\n_{C_3H_6}=0,15+0,075=0,225 mol\\n_{C_2H_6}=0,225 mol \end{cases}$
$\Rightarrow \begin{cases}\%H_2=25\% \\ \%C_3H_6=37,5\%\\\%C_2H_6=37,5\% \end{cases}$
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Trả lời 07-02-13 12:47 AM
viet130480
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