\(n_{CH3OH}=\frac{17,92}{32}=0,56\left(mol\right)\)
\(CH_3OH+CuO\rightarrow HCHO+Cu+H_2O\)
\(\rightarrow n_{HCHO}=n_{CH3OH}=0,56\left(mol\right)\)\(HCHO+4AgNO_3+6NH_3+2H_2O\rightarrow4Ag+\left(NH_4\right)_2CO_3+4NH_4NO_3\)
\(\rightarrow n_{Ag}=4n_{HCHO}=0,56.4=2,24\left(mol\right)\)
\(\rightarrow m_{Ag}=2,24.108=241,92\left(g\right)\)