\(C_6H_{12}O_6\left(0,03\right)+Ag_2O\xrightarrow[NH_3]{t^o}C_6H_{16}O_7+2Ag\left(0,06\right)\)
\(n_{C_6H_{12}O_6}=\dfrac{5,4}{180}=0,03\left(mol\right)\)
Vì \(H=95\%\)
\(\Rightarrow n_{Ag}=0,057\left(mol\right)\)
\(\Rightarrow m_{Ag}=6,156\left(g\right)\)