n Fe3O4=\(\dfrac{4,64}{232}=0,02mol\)
3Fe+2O2-to>Fe3O4
0,06----0,04---0,02
=>m O2=0,04.32=1,28g
\(n_{Fe_3O_4}=\dfrac{m}{M}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,06 0,02 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)
nFe3O4=4,64/232=0,02(mol)
PTHH: 3Fe+2O2 to⟶Fe3O4
0,06 mol ← 0,02 mol
mFe=56×0,06=3,36(g)
mO2=4,64–3,36=1,28(g)
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