\(6Fe+2O_2\underrightarrow{t^o}2Fe_3O_4\)
\(6mol\) \(2mol\) \(2mol\)
\(0,1mol\) \(\dfrac{1}{30}mol\) \(\dfrac{1}{30}mol\)
\(n_{Fe}=\dfrac{m}{M}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(n_{O_2}=\dfrac{V}{22,4}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
\(\text{Ta thấy }O_2\text{ dư,}Fe\text{ phản ứng hết}\)
\(m_{Fe_3O_4}=n.M=\dfrac{1}{30}.232\approx7,73\left(g\right)\)