Đáp án:

C. 123,35g

Giải thích các bước giải:

\(\begin{array}{l}
{n_Y} = 0,25mol\\
 \to {n_{{N_2}O}} + {n_{NO}} = 0,25mol\\
\dfrac{{{n_{{N_2}O}}}}{{{n_{NO}}}} = \dfrac{2}{3}\\
 \to {n_{{N_2}O}} = 0,1mol \to {n_{NO}} = 0,15mol
\end{array}\)

Bảo toàn electron ta có:

\(\begin{array}{l}
{n_{Zn}} = 0,65mol\\
Zn \to Z{n^{2 + }} + 2{\rm{e}}\\
2{N^{5 + }} + 8{\rm{e}} \to {N_2}^ + \\
{N^{5 + }} + 3{\rm{e}} \to {N^{2 + }}\\
2{n_{Zn}} = 8{n_{{N_2}O}} + 3{n_{NO}}\\
1,3mol > 1,25mol
\end{array}\)

Suy ra có tạo muối \(N{H_4}N{O_3}\)

\(\begin{array}{l}
Zn \to Z{n^{2 + }} + 2{\rm{e}}\\
2{N^{5 + }} + 8{\rm{e}} \to {N_2}^ + \\
{N^{5 + }} + 3{\rm{e}} \to {N^{2 + }}\\
2{N^{5 + }} + 8{\rm{e}} \to N{H_4}^ + \\
2{n_{Zn}} = 8{n_{{N_2}O}} + 3{n_{NO}} + 8{n_{N{H_4}^ + }}\\
 \to {n_{N{H_4}^ + }} = 0,00625mol\\
 \to {n_{Zn{{(N{O_3})}_2}}} = {n_{Zn}} = 0,65mol\\
 \to {n_{N{H_4}N{O_3}}} = {n_{N{H_4}^ + }} = 0,00625mol\\
 \to m = {m_{Zn{{(N{O_3})}_2}}} + {m_{N{H_4}N{O_3}}}\\
 \to m = 0,65 \times 189 + 0,00625 \times 80 = 123,35g
\end{array}\)