1) \(n_{Al_2\left(SO_4\right)_3}=\dfrac{68,4}{342}=0,2\left(mol\right)\)
Ta có: \(n_{Al}=2n_{Al_2\left(SO_4\right)_3}=2\times0,2=0,4\left(mol\right)\)
Ta có: \(n_S=3n_{Al_2\left(SO_4\right)_3}=3\times0,2=0,6\left(mol\right)\)
Ta có: \(n_O=12n_{Al_2\left(SO_4\right)_3}=12\times0,2=2,4\left(mol\right)\)
2) a) \(n_{Fe_3O_4}=\dfrac{69,6}{232}=0,3\left(mol\right)\)
Ta có: \(n_{Fe}=3n_{Fe_3O_4}=3\times0,3=0,9\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,9\times56=50,4\left(g\right)\)
Ta có: \(n_O=4n_{Fe_3O_4}=4\times0,3=1,2\left(mol\right)\)
\(\Rightarrow m_O=1,2\times16=19,2\left(g\right)\)
b) \(n_O=\dfrac{2,4\times10^{23}}{6\times10^{23}}=0,4\left(mol\right)\)
\(\Rightarrow m_O=0,4\times16=6,4\left(g\right)\)
