Đáp án:
C
Giải thích các bước giải:
\(\begin{array}{l}
{n_{{H_2}O}} = \dfrac{2}{{18}} = \dfrac{1}{9}mol\\
{n_H} = 2{n_{{H_2}O}} = \dfrac{2}{9}mol\\
Bt\,C:\\
{m_C} = 2 - \dfrac{2}{9} \times 1 = \frac{{16}}{9}g\\
{n_C} = \dfrac{{16}}{9}:12 = \frac{4}{{27}}mol\\
CTPT\,A:{C_x}{H_y}\\
{n_C}:{n_H} = \dfrac{4}{{27}}:\frac{2}{9} = 2:3\\
\Rightarrow CTDGN:{C_2}{H_3}\\
\Rightarrow CTPT:{C_4}{H_6}\\
{C_4}{H_6} + AgN{O_3} + N{H_3} \to N{H_4}N{O_3} + {C_4}{H_5}Ag\\
{n_{{C_4}{H_6}}} = \dfrac{{2,7}}{{54}} = 0,05mol\\
{n_{{C_4}{H_5}Ag}} = {n_{{C_4}{H_6}}} = 0,05mol\\
m = 0,05 \times 161 = 8,05g
\end{array}\)
