a) \(\left\{{}\begin{matrix}n_{Fe}=\dfrac{m}{M}=\dfrac{28}{56}=0,5\left(mol\right)\\n_{H_2SO_4}=\dfrac{m}{M}=\dfrac{9,8}{98}=0,1\left(mol\right)\\n_{H_2}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{1,2.10^{23}}{6.10^{23}}=0,2\left(mol\right)\\n_{CO_2}=\dfrac{V_{\left(đktc\right)}}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}n_{N_2}=\dfrac{m}{M}=\dfrac{5,6}{28}=0,2\left(mol\right)\\n_{SO_2}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1\left(mol\right)\\n_{CO}=\dfrac{\text{Số phân tử}}{6.10^{23}}=\dfrac{1,8.10^{23}}{6.10^{23}}=0,3\left(mol\right)\end{matrix}\right.\)
`=> n_{hh} = n_{N_2} + n_{SO_2} + n_{CO} = 0,2 + 0,1 + 0,3 = 0,6 (mol)`
`=> V_{hh} = n.22,4 = 0,6.22,4 = 13,44 (l)`
