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$*)m gam X:\begin{cases}RCOOR_1:x mol \\ R_2COOH:y mol \end{cases}+0,15 mol NaOH$
$\Rightarrow \begin{cases}x+y\leqslant 0,15 mol \\ n_{R_1OH}=x mol\\m_{R_1OH}=3,36 gam \end{cases}$
$*)0,5m gam X+NaHCO_3\rightarrow CO_2$
$\Rightarrow 0,5y=\frac{0,54}{22,4}\Rightarrow y=0,0482 mol\Rightarrow x<0,1018\Rightarrow M_{R_1OH}>\frac{3,36}{0,1018}\Rightarrow M_{R_1}>16$
$\Rightarrow R_1:C_2H_5\Rightarrow x=\frac{3,36}{46} mol$
$\Rightarrow \frac{3,36}{46}.(R+44+23)+\frac{0,54}{11,2}.(R_2+44+23)=12,84\Rightarrow R+0,66R_2=64,56$
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Trả lời 24-01-13 02:30 AM
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