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 35L 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

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