$a,\\Đặt \begin{cases}n_{CH_3CH_2COOH}=x(mol)\\n_{HCOOH}=y(mol)\end{cases}\\ PTHH:\\ CH_3CH_2COOH+NaOH\to CH_3CH_2COONa+H_2O\\ HCOOH+NaOH\to HCOONa+H_2O\\ \Rightarrow \begin{cases}74x+46y=19,4\\96x+68y=26\end{cases}\Rightarrow\begin{cases}x=0,2\\y=0,1\end{cases}\\ \Rightarrow \%m_{CH_3CH_2COOH}=\dfrac{0,2.74}{19,4}.100\%=76,29\%\\ \Rightarrow \%m_{HCOOH}=100\%-76,29\%=23,71\%\\ b,\\ HCOOH+Ag_2O\xrightarrow{NH_3} H_2O+CO_2+2Ag\downarrow\\ n_{Ag}=2.n_{HCOOH}=0,2(mol)\\ \Rightarrow m_{Ag}=0,2.108=21,6(g)$