Question

Để hòa tan 4g oxit FexOy cần dùng 52,14ml dd HCl 10% ( d= 1,05 g/ml). Xác định CTHH của oxit sắt

Answer 1

PTHH: Fe_xO_y + 2yHCl ----> xFeCl_2y/x + yH₂O

=> m_\(ddHCl\) = 1,05.52,14 = 54,747 (g)

=> m_\(HCl\) = \(\dfrac{54,747.10\%}{100\%}=5,4747\left(g\right)\)

=> n_\(HCl\) = \(\dfrac{5,4747}{36,5}=0,15\left(mol\right)\)

Theo PTHH: n_{\(Fe_xO_y\)} = \(\dfrac{1}{2y}\)n_\(HCl\) = \(\dfrac{0,15}{2y}\left(mol\right)\)

=> M_{\(Fe_xO_y\)} = \(\dfrac{4}{\dfrac{0,15}{2y}}=\dfrac{8y}{0,15}\)

<=> 56x + 16y = \(\dfrac{8y}{0,15}\)

<=> \(0,15\left(56x+16y\right)=8y\)

<=> 8,4x = 5,6y

=> \(\dfrac{x}{y}=\dfrac{2}{3}\) => x = 2, y = 3

=> CTHH: Fe₂O₃

Answer 2

PTHH: \(Fe_xO_y+2yHCl\rightarrow xFeCl_{\dfrac{2y}{x}}+yH_2O\)

Ta có: \(m=D.V=1,05.52,14=54,747g\)

=> \(n_{HCl}=\dfrac{54,747.10\%}{36,5}\approx0,15mol\)

Cứ 1 mol Fe_xO_y --> 2y mol HCl

56x + 16y (g) --> 2y mol

4 (g) --> 0,15 mol

=> \(8,4x+2,4y=8y\)

=> \(8,4x=5,6y\)

=> \(\dfrac{x}{y}=\dfrac{5,6}{8,4}=\dfrac{2}{3}\)

=> CT của oxit sắt cần tìm là Fe₂O₃