Khí màu nâu là NO2
\(Fe\rightarrow Fe^{3+}+3e\) \(\) \(2H^++NO_3^-+1e\rightarrow NO_2+H_2O\)
Bảo toàn e: \(n_{Fe}.3=n_{NO_2}.1\\ \Rightarrow n_{NO_2}=0,3\left(mol\right)\)
Ta có : \(n_{H^+}=n_{HNO_3}=2n_{NO_2}=0,6\left(mol\right)\\ \Rightarrow V_{HNO_3}=\dfrac{0,6}{0,5}=1,2\left(l\right)\)
Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)
\(PTHH:Fe+2HNO_3--->Fe\left(NO_3\right)_2+H_2\uparrow\)
Theo PT: \(n_{HNO_3}=2.n_{Fe}=2.0,1=0,2\left(mol\right)\)
\(\Rightarrow V_{dd_{HNO_3}}=\dfrac{0,2}{0,5}=0,4\left(lít\right)\)