Question

Answer 1

Đáp án C

Ta có :

\(\dfrac{m_R}{m_O} = \dfrac{2R}{16.3} = \dfrac{11,2}{4,8}\\ \Rightarrow R = 56(Fe)\)

Vậy X là Fe₂O₃(Sắt III oxit)

Answer 2

Ta có : \(\dfrac{m_O}{m_R}=\dfrac{M_O.3n_X}{M_R.2n_X}=\dfrac{4,8}{11,2}=\dfrac{3}{7}=\dfrac{3M_O}{2M_R}\)

\(\Rightarrow M_R=\dfrac{7}{3}.\dfrac{3}{2}.M_O=56\)

=> R là sắt ( Fe )

=> Đáp án C .