\(n_{H_2}=0,3\left(mol\right);n_{O_2}=0,75\left(mol\right);n_{N_2}=0,8\left(mol\right);n_{CO_2}=0,15\left(mol\right)\)a) \(\Sigma n=0,3+0,75+0,8+0,15=2\left(mol\right)\)\(\%n_{H_2}=\dfrac{0,3}{2}.100=15\%\)\(\%n_{O_2}=\dfrac{0,75}{2}.100=37,5\%\)\(\%n_{N_2}=\dfrac{0,8}{2}.100=40\%\)\(\%n_{CO_2}=\dfrac{0,15}{2}.100=7,5\%\)b) \(\overline{M}_X=\dfrac{m_X}{n_X}=\dfrac{0,3.2+0,75.32+0,8.28+0,15.44}{2}=26,8\)(g/mol)dX/SO2=\(\dfrac{26,8}{64}=0,41875\)dX/H2S=\(\dfrac{26,8}{34}=0,788\)dX/NO2=\(\dfrac{26,8}{46}=0,582\)dX/NO=\(\dfrac{26,8}{30}=0,893\) Câu trả lời: \(n_{H_2}=0,3\left(mol\right);n_{O_2}=0,75\left(mol\right);n_{N_2}=0,8\left(mol\right);n_{CO_2}=0,15\left(mol\right)\)a) \(\Sigma n=0,3+0,75+0,8+0,15=2\left(mol\right)\)\(\%n_{H_2}=\dfrac{0,3}{2}.100=15\%\)\(\%n_{O_2}=\dfrac{0,75}{2}.100=37,5\%\)\(\%n_{N_2}=\dfrac{0,8}{2}.100=40\%\)\(\%n_{CO_2}=\dfrac{0,15}{2}.100=7,5\%\)b) \(\overline{M}_X=\dfrac{m_X}{n_X}=\dfrac{0,3.2+0,75.32+0,8.28+0,15.44}{2}=26,8\)(g/mol)dX/SO2=\(\dfrac{26,8}{64}=0,41875\)dX/H2S=\(\dfrac{26,8}{34}=0,788\)dX/NO2=\(\dfrac{26,8}{46}=0,582\)dX/NO=\(\dfrac{26,8}{30}=0,893\)