\(n_{NaOH}=0,2.0,1=0,02\left(mol\right)\)
=> \(n_{HCOOCH_3}+n_{HCOOC_6H_5}=0,02\left(mol\right)\) => nHCHO = 0,02 (mol)
Ta có
\(HCOOCH_3\rightarrow2Ag\)
\(HCOOC_6H_5\rightarrow2Ag\)
\(HCHO\rightarrow4Ag\)
=> nAg = 0,02.2 + 0,02.4 = 0,12 (mol)
=> mAg = 0,12.108 = 12,96 (g)
=> C