Question

Answer 1

Ta có: \(M_{X_2O_3}=80.2=160\left(g\right)\)

Mà: \(M_{X_2O_3}=NTK_X.2+16.3=160\left(g\right)\)

\(\Leftrightarrow NTK_X=56\left(đvC\right)\)

Vậy X là nguyên tố sắt (Fe)

Answer 2

biết \(PTK_{H_2}=2.1=2\left(đvC\right)\)

vậy \(PTK_{hợpchất}=2.80=160\left(đvC\right)\)

ta có:

\(2X+3O=160\)

\(2X+3.16=160\)

\(2X+48=160\)

\(2X=160-48\) \(=112\)

\(X=\dfrac{112}{2}=56\left(đvC\right)\)

\(\Rightarrow X\) là \(Fe\) (sắt)