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`
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: \(Fe_2O_3\)