\(\begin{array}{l}
M_A=22.2=44(g/mol)\\
TH_1:\,\rm A\,là\,N_2\,\to M_A=28\,(loại)\\
TH_2:\,A:N_xO_y\\
\to 14x+16y=44\\
\to 7x+8y=22\\
\to (x;y)=(2;1)\\
\to A:N_2O\\
n_{N_2O}=\frac{4,4}{44}=0,1(mol)\\
Zn^0\to Zn^{+2}+2e\\
2N^{+5}+8e\to 2N^{+1}\\
Bao\,toan\,e:\,2n_{Zn}=8n_{N_2O}\\
\to n_{Zn}=0,4(mol)\\
\to m=0,4.65=26(g)
\end{array}\)