Phần 1:
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
\(\Rightarrow n_{CH3COOH}=n_{NaOH}=0,1\left(mol\right)\)
Phần 2:
\(HCHO+4AgNO_3+6NH_3+2H_2O\rightarrow4Ag+\left(NH_4\right)_2CO_3+4NH_4NO_3\)
Ta có:
\(n_{Ag}=\frac{32,4}{108}=0,3\left(mol\right)\)
\(\Rightarrow n_{HCHO}=\frac{1}{4}n_{Ag}=0,075\left(mol\right)\)
\(\Rightarrow m=0,2.60+0,15.30=16,5\left(g\right)\)
\(m_{CH3COOH}=0,2.60=12\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH3COOH}=\frac{12}{16,5}.100\%=72,7\%\\\%m_{HCHO}=100\%-72,7\%=27,3\%\end{matrix}\right.\)