Asample 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

The answer is celsius