Question

Answer 1

Câu 21:

CTHH: Fe_xO_y

Có: \(\dfrac{m_{Fe}}{m_O}=\dfrac{7}{3}\Rightarrow\dfrac{56x}{16y}=\dfrac{7}{3}\Rightarrow\dfrac{x}{y}=\dfrac{2}{3}\)

=> CTHH: Fe₂O₃

Câu 22:

a)

\(M_Y=0,5.32=16\left(g/mol\right)\)

\(M_X=16.2,125=34\left(g/mol\right)\)

b)

- Xét X:

Có: \(n_H:n_S=\dfrac{5,88\%}{1}:\dfrac{94,12\%}{32}=2:1\)

=> CTHH: (H₂S)_n

Mà M_X = 34 (g/mol)

=> n = 1 => CTHH: H₂S

- Xét Y:

Có: \(n_C:n_H=\dfrac{75\%}{12}:\dfrac{25\%}{1}=1:4\)

=> CTHH: (CH₄)_m

Mà M_Y =  16 (g/mol)

=> m = 1 => CTHH: CH₄