\(n_{Mg}=\dfrac{19,2}{24}=0,8\left(mol\right)\); \(n_{N_xO_y}=\dfrac{35,84}{22,4}=1,6\left(mol\right)\)
\(Mg^0-2e\rightarrow Mg^{+2}\)
0,8-->1,6
\(xN^{+5}+\left(5x-2y\right)e\rightarrow N_x^{+\dfrac{2y}{x}}\)
1,6(5x-2y) <---1,6
Bảo toàn e: 1,6(5x-2y) = 1,6
=> 5x - 2y = 1
=> x = 1; y = 2 thỏa mãn
CTHH: NO2
=> C