$n_{O_2}=20\%n_{kk}=20%.\dfrac{77,28}{22,4}=0,69(mol)$
$n_{H_2O}=\dfrac{6,48}{18}=0,36(mol)$
Bảo toàn $O$:
$2n_{O_2}=2n_{CO_2}+n_{H_2O}$
$\to n_{CO_2}=\dfrac{0,69.2-0,36}{2}=0,51(mol)$
$\to m_{hh}=m_C+m_H=0,51.12+0,36.2.1=6,84g\ne 4,84g$ (bạn xem lại đề)
* Nếu bỏ $4,84g$ thì chọn $C$