\({n_{Ag}} = \frac{{21,6}}{{108}} = 0,2\,(mol)\)
CH3CH=O + 2AgNO3 + 3NH3 → CH3COONH4 + 2Ag↓ + 2NH4NO3
nCH3CHO = ½ nAg = 0,1 (mol)
\(C{\% _{C{H_3}CHO}} = \frac{{{m_{C{H_3}CHO}}}}{{m{\,_{dd\,C{H_3}CHO}}}}.100\% \, = \frac{{0,1.44}}{{50}}.100\% \)
\(= 8,8\% \)