Giải thích các bước giải:
PTHH $R_xO_y + yH_2 → xR + yH_2O$
$n_{H_2}=$ $\dfrac{0,336}{22,4}=0,015(mol)$
Theo PTHH $n_{O(oxit)}=n_{H_2}=0,015(mol) \to m_{O(oxit)}=0,015*16=0,24(g)$
$\to m_R=1,16-0,24=0,92(g)\to n_R=$ $\dfrac{0,92}{M_R}(mol)$
Theo PTHH: $n_R=$ $\dfrac{x}{y}*n_{H_2}=$ $\dfrac{0,015x}{y}(mol)$
$→$ $\dfrac{0,92}{M_R}=$ $\dfrac{0,015x}{y}→M_R=$ $\dfrac{2y}{x}*$ $\dfrac{92}{3}??$