a sample of methane gas having a volume of 2.8 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 3 5L at 31°C and 1.25 atm. The mixture was ignited to form carbon dioxide and water. Calculate the volume of carbon dioxide formed at a pressure of 2.5 atm and a temperature of 125°C
1 answer:
When the balanced reaction equation for the reaction of methane gas and O2 is: CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) we will use the ideal gas equation to get the moles of each gas: 1) moles of methane: PV= n RT when p is the pressure =1.65 atm V is the volume = 2.8 L R is the ideal gas constant = 0.0821 T is the temperature in Kelvin = 25+ 273= 298 K ∴n of CH4 = 1.65 * 2.8 / 0.0821 * 298 K = 0.189 mole 2) moles of O2: PV = nRT when P = 1.25 atm and V= 35L and R = 0.0821 L atm / mol K T = 31 + 273 = 304 K so, by substitution: n of O2= 1.25 atm * 35 L / 0.0821 * 304 = 1.75 mol from the balanced reaction equation, we can see that the molar ratio between O2 and CH4 is 2: 1 ∴ moles of O2 = 2 * .0189 moles of methane = 0.378 mole so the O2 exists in excess and CH4 is the limiting reactant when 1 mole of CH4 → 1 mole ofCO2 1 mol of CO2 → 22.4 L CO2 volume of CO2 at STP =( 0.189 molesCH4 /1)*(1mol CO2)/(1 mole CH4)*(22.4L CO2)/(1molCO2) = 4.23 L then we will use this formula to get the correct value of V of CO2 at laboratory conditions: P1V1 /T1 = P2V2/T2 when at STP conditions: P1= 1 atm V1 = 4.23 L T1 = 273 and the laboratories conditions are: P2 = 2.5 atm T2 = 125 + 273 = 398 K ∴ V2 = 1 atm * 4.23 L *398 / 273 * 2.5 atm ∴ V2 = 15.4 L ∴ the volume of CO2 formed at pressure of 2.5 atm and T = 125°C is 15.4 L
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