Question

Cho 8,4 gam Fe tác dụng vừa đủ với x lít khí oxi (đktc) sau phản ứng thu được y gam Fe₃O₄. Tính x, y ( Fe: 56, O:16)

Answer 1

Ta có: \(n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\)

\(pthh:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

          0,15 -> 0,1   ------->  0,05        (mol)

\(\Rightarrow\left\{{}\begin{matrix}V_{O_2}=0,1.22,4=2,24\left(lít\right)\\m_{Fe_3O_4}=0,05.232=11,6\left(g\right)\end{matrix}\right.\)

Answer 2

\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\\ n_{Fe}=\dfrac{8,4}{56}=0,15\left(mol\right)\\ n_{O_2}=\dfrac{2.n_{Fe}}{3}=\dfrac{2.0,15}{3}=0,1\left(mol\right)\\ n_{Fe_3O_4}=\dfrac{n_{Fe}}{3}=\dfrac{0,15}{3}=0,05\left(mol\right)\\ \Rightarrow x=22,4.0,1=2,24\left(l\right)\\ y=232.0,05=11,6\left(g\right)\)

Answer 3

PTHH : 3Fe + 2O₂ -----> Fe₃O₄

            0,15.....0,1............0,05

n_Fe = \(\dfrac{8,4}{56}=0,15\left(mol\right)\)

n_o2 = \(\dfrac{3}{2}.0,15=0,1\left(mol\right)\)

=> V_O2 = 0,1 . 22,4 = 2,24 (l)

n_Fe3O4 = 0,05 ( mol )

=> m_Fe3O4 = 0,05 . ( 56x3 + 16x4 ) = 11,6 (g)