$\Delta m_{dd}=3,78g=m_M-m_{\text{khí}}$
$\Rightarrow m_{\text{khí}}=5,2-3,78=1,42g$
Gọi x, y là số mol $NO$, $N_2O$
$\Rightarrow 30x+44y=1,42$ (1)
$n_{\text{khí}}=\dfrac{1,008}{22,4}=0,045(mol)$
$\Rightarrow x+y=0,045$ (2)
$(1)(2)\Rightarrow x=0,04; y=0,005$
$3M+4nHNO_3\to 3M(NO_3)_n+nNO+2nH_2O$
$8M+10nHNO_3\to 8M(NO_3)_n+nN_2O+5nH_2O$
$\Rightarrow n_M=\dfrac{3.0,04}{n}+\dfrac{0,005.8}{n}=\dfrac{0,16}{n}(mol)$
$\Rightarrow M_M=\dfrac{5,2n}{0,16}=\dfrac{65n}{2}$
$n=2\Rightarrow M_M=65(Zn)$
Vậy M là kẽm.