\({n_{Ag}} = \frac{{21,6}}{{108}} = 0,2\,(mol)\)
a) PTHH:
CH3CHO + 2AgNO3 + 3NH3 → CH3COONH4 +2Ag↓ + 2NH4NO3 (1)
CH3COOH + NaOH → CH3COONa + H2O (2)
b) \({n_{C{H_3}CHO}} = \dfrac{1}{2}{n_{Ag}} = \dfrac{{0,2}}{2}\, = 0,1(mol)\)
\(\begin{gathered}
\,\,\,\,\,\,{n_{C{H_3}CHO}} = \dfrac{1}{2}{n_{Ag}} = \dfrac{{0,2}}{2}\, = 0,1(mol) \hfill \\
= > {m_{C{H_3}CHO}} = 0,1.44 = 4,4\,(g) \hfill \\
\% C{H_3}CHO = \dfrac{{4,4}}{{10}}.100\% = 44\% \hfill \\
= > \% C{H_3}COOH = 100\% - 44\% = 66\% \hfill \\
\,{m_{C{H_3}COOH}} = 10 - 4,4 = 6,6\,(g) \hfill \\
= > \,{n_{C{H_3}COOH}} = \dfrac{{6,6}}{{60}} = 0,11\,(mol) \hfill \\
Theo\,(2)\,:{n_{NaOH}} = {m_{C{H_3}COOH}} = 0,11\,(mol) \hfill \\
= > {V_{NaOH}} = \frac{{0,11}}{{0,2}} = 0,55\,(M) \hfill \\
\end{gathered} \)