Bạn tham khảo:
$4A+nO_2 \xrightarrow{t^{o}} 2A_2O_n$
Cách 1:
BTKL:
$m_A+m_{O_2}=m_{A_2O_n}$
$6,4+m_{O_2}=8$
$m_{O_2}=1,6(g)$
$\to n_{O_2}=0,05(mol)$
$n_A=\frac{0,2}{n}(mol)$
$M_A=\frac{6,4.n}{0,2}=32.n$
$n=2; A=64$
$ \to Cu$
Cách 2:
$n_A=a(mol)$
$\to n_{A_2O_n}=0,5a(mol)$
$\frac{a.A}{0,5a.(2A+16n)}=\frac{6,4}{8}$
$\to \frac{A}{0,5(.2A+16n)}=\frac{6,4}{8}$
$A=32.n$
$n=2; A=64$
$ \to Cu$