Bài 1:
Ta có: \(m_{Fe_3O_4}=\dfrac{69,6}{232}=0,3\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,3.3.56=50,4\left(g\right)\)
\(m_O=0,3.4.16=19,2\left(g\right)\)
Bài 2:
Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)
\(\Rightarrow n_{Fe_3O_4}=\dfrac{1}{3}n_{Fe}=0,1\left(mol\right)\)
\(\Rightarrow m_{Fe_3O_4}=0,1.232=23,2\left(g\right)\)