Đáp án:
${m_{Ag}} = 21,6\,\,gam$
Giải thích các bước giải:
${n_{C{H_3}CHO}} = \dfrac{{4,4}}{{44}} = 0,1\,\,mol$
$\begin{gathered} C{H_3}CHO\xrightarrow{{AgN{O_3}/N{H_3}}}2Ag\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \hfill \\ 0,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,mol \hfill \\ \end{gathered} $
$\to {m_{Ag}} = 0,2.108 = 21,6\,\,gam$