Question

Dùng 3,36 lít \(H_2\) (đktc) để khử hoàn toàn 8g \(Fe_xO_y\)

a) Viết phương trình

b) Tìm CTHH của \(Fe_xO_y\)

Answer 1

a) \(n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

PTHH: \(Fe_xO_y+yH_2\xrightarrow[]{t^o}xFe+yH_2O\)

b) Theo PT: \(n_{Fe_xO_y}=\dfrac{1}{y}n_{H_2}=\dfrac{0,15}{y}\left(mol\right)\)

`=>` \(M_{Fe_xO_y}=\dfrac{8}{\dfrac{0,15}{y}}=\dfrac{160}{3}y\left(g/mol\right)\)

`=> 56x + 16y = (160y)/3`

`=> 56x = (112y)/3`

`=> x/y = (112)/3 : 56 = 2/3`

`=> Fe_xO_y: Fe_2O_3`

Answer 2

a)   \(yH_2+Fe_xO_y->yH_2O+xFe\)

PT: y---------1-------------------------------          (mol)

ĐB:\(0,15\)-\(\dfrac{0,15}{y}\)--------------------------              (mol)

\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

Ta có: \(m_{Fe_xO_y}=n.M\)

               \(8=\dfrac{0,15}{y}.\left(56x+16y\right)\)

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

               \(8y=8,4x+2,4y\)

              \(8y-2,4y=8,4x\)

              \(5,6y=8,4x\)

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

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

\(=>x=2;y=3\)

CTHH: \(Fe_2O_3\)