\(n_{C_{12}H_{22}O_{11}}=\dfrac{3.42}{342}=0.01\left(mol\right)\)
\(\Rightarrow n_{pư}=0.01\cdot80\%=0.008\left(mol\right)\)
\(n_{C_6H_{12}O_6}=0.008\cdot2=0.016\left(mol\right)\)
\(\Rightarrow n_{Ag}=2\cdot0.016=0.032\left(mol\right)\)
\(m=3.456\left(g\right)\)
\(m_{Ag}=0.008\cdot108=0.864\left(g\right)\)
