The partial pressure of \(\mathrm{CH}_{4}(g)\) is \(0.175 \mathrm{~atm}\) and that of \(\mathrm{O}_{2}(g)\) is \(0.250\) atm in a mixture of the two gases. a. What is the mole fraction of each gas in the mixture? b. If the mixture occupies a volume of \(10.5 \mathrm{~L}\) at \(65^{\circ} \mathrm{C}\), calculate the total number of moles of gas in the mixture. c. Calculate the number of grams of each gas in the mixture.

a. Calculate the mole fraction of each gas: Step 1: Total pressure, \(P_{total} = 0.175 + 0.250 = 0.425\) atm Step 2: Mole fractions: \(X_{CH4} = \frac{0.175}{0.425} = 0.412\) and \(X_{O2} = \frac{0.250}{0.425} = 0.588\) b. Calculate the total number of moles of gas in the mixture: Step 3: \(T (K) = 65 ^\circ C + 273.15 = 338.15 K\) Step 4: \(n = \frac{PV}{RT} = \frac{(0.425 \mathrm{~atm})(10.5 \mathrm{~L})}{(0.0821 \frac{\mathrm{L\cdot atm}}{\mathrm{mol\cdot K}})(338.15 \mathrm{~K})} = 0.179\) mol c. Calculate the number of grams of each gas in the mixture: Step 5: Moles of CH4 = \(0.412 * 0.179 = 0.0737\) mol Moles of O2 = \(0.588 * 0.179 = 0.105\) mol Step 6: Mass of CH4 (grams) = \(0.0737\,\mathrm{mol} * 16.04 \frac{\mathrm{g}}{\mathrm{mol}} = 1.18 \mathrm{~g}\) Mass of O2 (grams) = \(0.105\,\mathrm{mol} * 32.00 \frac{\mathrm{g}}{\mathrm{mol}} = 3.36 \mathrm{~g}\)