A \(1.604-g\) sample of methane \(\left(\mathrm{CH}_{4}\right)\) gas and \(6.400 \mathrm{~g}\) oxygen gas are sealed into a 2.50-L vessel at \(411^{\circ} \mathrm{C}\) and are allowed to reach equilibrium. Methane can react with oxygen to form gaseous carbon dioxide and water vapor, or methane can react with oxygen to form gaseous carbon monoxide and water vapor. At equilibrium, the pressure of oxygen is \(0.326 \mathrm{~atm}\), and the pressure of water vapor is \(4.45\) atm. Calculate the pressures of carbon monoxide and carbon dioxide present at equilibrium.

First, we need to convert the given masses of methane (CH4) and oxygen (O2) to moles using their respective molar masses. For methane: \[Molar~mass~of~CH_4 = 12.01 + 4(1.01) = 16.05~g/mol\] Moles of CH4 = \(\frac{1.604~g}{16.05~g/mol} = 0.100~mol\) For oxygen: \[Molar~mass~of~O_2 = 2(16.00) = 32.00~g/mol\] Moles of O2 = \(\frac{6.400~g}{32.00~g/mol} = 0.200~mol \)

Now, we will convert the moles of methane and oxygen to their initial pressures using the Ideal Gas Law equation: \(PV=nRT\), and the given temperature and volume of the vessel. Temperature: \[ T = 411 ^{\circ}C + 273.15 = 684.15~K \] For methane: \[P_{CH_4} = \frac{(0.100~mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(684.15 K)}{2.50 L} = 2.23~atm \] For oxygen: \[P_{O_2} = \frac{(0.200~mol)(0.0821 \frac{L \cdot atm}{mol \cdot K})(684.15 K)}{2.50 L} = 4.46~atm \]

Now, we will write the balanced chemical reactions that can occur. Reaction 1: Methane and oxygen produce carbon dioxide and water \[CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O \] Reaction 2: Methane and oxygen produce carbon monoxide and water \[2CH_4 + 3O_2 \rightarrow 2CO + 4H_2O \]

We can determine the decrease in CH4 and O2 pressures as CH4 reacts with O2 in both reactions. Then we can find the increase in CO and CO2 pressures at equilibrium. \(P_{O_2}\) decreases by 4.46 atm - 0.326 atm = 4.134 atm Decrease in \(P_{CH_4}\) can be split between two reactions: \(x\) atm due to reaction 1 and \(2y\) atm due to reaction 2, i.e., \(x + 2y = 2.23\) atm For reaction 1: \(\frac{x}{2} = \frac{4.134}{2}\), solving for x, we get \(x = 4.134\) atm For reaction 2: \(\frac{y}{3} = 4.134\), solving for y, we get \(y = 0.096\) atm Increased pressure for \(P_{CO_2}\) = 4.134 atm (due to reaction 1) Increased pressure for \(P_{CO}\) = 0.096 atm (due to reaction 2) Finally, the pressures at equilibrium for carbon monoxide (CO) and carbon dioxide (CO2) are 0.096 atm and 4.134 atm, respectively.