\(n_{Fe_3O_4}=\dfrac{m_{Fe_3O_4}}{M_{Fe_3O_4}}=\dfrac{4,64}{232}=0,02mol\)
\(3Fe+2O_2\rightarrow\left(t^o\right)Fe_3O_4\)
0,06 0,04 0,02 ( mol )
\(m_{Fe}=n_{Fe}.M_{Fe}=0,06.56=3,36g\)
\(m_{O_2}=n_{O_2}.M_{O_2}=0,04.32=1,28g\)
\(pthh:3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)
Ta có: \(n_{Fe_3O_4}=\dfrac{4,64}{232}=0,02\left(mol\right)\)
Theo pt: \(n_{Fe}=3.0,02=0,06\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,06.56=3,36\left(g\right)\)
Theo pt: \(n_{O_2}=2.0,02=0,04\left(mol\right)\)
\(\Rightarrow m_{O_2}=0,04.32=1,28\left(g\right)\)