\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
\(0.2..................................................0.1\)
\(n_{Ag}=\dfrac{21.6}{108}=0.2\left(mol\right)\)
\(CH_3CHO+2AgNO_3+3NH_3+H_2O\rightarrow CH_3COONH_4+2Ag+2NH_4NO_3\)
\(0.1........................................................................................0.2\)
\(n_X=0.2+0.1=0.3\left(mol\right)\)
\(\%n_{C_2H_5OH}=\dfrac{0.2}{0.3}\cdot100\%=66.67\%\)
\(\%n_{CH_3CHO}=33.33\%\)