Gọi CTHH oxit là $M_2O_n$
Muối tạo thành là $M(NO_3)_n$
$n_{M_2O_n}=\dfrac{2,8}{2M+16n}(mol)$
$\Rightarrow n_{M(NO_3)_n}=\dfrac{5,6}{2M+16n}(mol)$
Ta có: $\dfrac{5,6}{2M+16n}=\dfrac{10,26}{M+62n}$
$\Leftrightarrow 10,26(2M+16n)=5,6(M+62n)$
$\Leftrightarrow M=12n$
$n=2\to M=24(Mg)$