\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ n_{O_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(PTHH:3Fe+2O_2-^{t^o}>Fe_3O_4\)
tỉ lệ: 3 : 2 : 1
n(mol) 0,1: 0,5
n(mol p/u) 0,1--->`1/15`-->`1/30`
\(\dfrac{n_{Fe}}{3}< \dfrac{n_{O_2}}{2}\left(\dfrac{0,1}{3}< \dfrac{0,5}{2}\right)\)
`=>` `Fe` hết , `O_2` dư
`=>` tính theo `Fe`
\(m_{Fe_3O_4}=n\cdot M=\dfrac{1}{30}\cdot\left(56\cdot3+16\cdot4\right)\approx7,73\left(g\right)\)
\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right);n_{O_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
PTHH: \(3Fe+2O_2\xrightarrow[]{t^o}Fe_3O_4\)
Ban đầu: 0,1 0,5
Sau pư: 0 \(\dfrac{13}{30}\) \(\dfrac{1}{30}\)
\(\Rightarrow m_{Fe_3O_4}=\dfrac{1}{30}.232=\dfrac{116}{15}\left(g\right)\)
