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$\begin{cases}n_H=2n_{H_2O}=0,15 mol \\ n_C=\frac{2,65}{106}+\frac{1,68}{22,4}=0,1 mol\\n_{Na}=2.
\frac{2,65}{106}=0,05 mol \end{cases}\Rightarrow m_O=4,1-0,15-1,2-1,15=1,6\Rightarrow n_O=0,1$
$\Rightarrow n_C:n_H:n_O:n_{Na}=0,1:0,15:0,1:0,05=2:3:2:1\Rightarrow CTDG:C_2H_3O_2Na$
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Trả lời 20-01-13 01:27 AM
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