\(n_{Fe}=0,13\left(mol\right);n_{NO}=0,1\left(mol\right)\)
\(Fe+4HNO_3\rightarrow Fe\left(NO_3\right)_3+NO+2H_2O\)
\(\Rightarrow n_{Fe_{pư}}=n_{Fe\left(NO3\right)3}=n_{NO}=0,1\left(mol\right)\)
Dư 0,03 mol Fe
\(Fe+2Fe\left(NO_3\right)_3\rightarrow3Fe\left(NO_3\right)_2\)
\(\Rightarrow\) Dư 0,04 mol Fe(NO3)3. Tạo 0,09 mol Fe(NO3)2.
\(n_{Fe^{2+}}=0,09\left(mol\right)\)
\(n_{NO3^-}=3n_{Fe\left(NO3\right)3}+2n_{Fe\left(NO3\right)2}=0,3\left(mol\right)\)
\(3Fe^{2+}+4H^++NO_3^-\rightarrow3Fe^{3+}+NO+2H_2O\)
\(\Rightarrow Fe^{2+}\) hết , \(NO_3^-\) dư. Cần 0,12 mol \(H^+\)
\(\Rightarrow n_{HCl}=0,12\left(mol\right)\)
