Ethylene gas burns in air according to equation C2H4(g) + 3O2(g) 2CO2(g)+2H2(l) If 13.8 L of C2H4 at 21 ˚ C and 1.083 atm burns completely in oxygen, calculate the volume of CO2 produced, assuming the CO2 is measured at 44 ˚ C at 0.989 atm?

Treating the gases involved in this reaction as ideal gases means we can solve this problem using the Ideal Gas Law #pV = nRT#
Where #p# = pressure (atm), #V# = volume (L), #n# = number of moles, #T# = temperature (K), and #R# = the gas constant (#8.314 JK^-1mol^-1#)

First calculate the number of moles of ethylene gas we are burning by rearranging the ideal gas law for n:
#n = (pV) / (RT)#

We need to convert temperature to Kelvin , and then we can just substitute in the values.

To convert Celsius to Kelvin, you add 273.15 to the value

#21 ^@C# = #294.15 K#

#n = (13.8L xx 1.083) / (8.413 xx 294.15)#

#n = 6.11 xx 10^-3# moles

Looking at the balanced equation, ethylene gas is in a 1 : 2 molar ratio with #CO_2# so we can calculate the number of moles of #CO_2# produced when ethylene burns completely by multiplying the no. moles of ethylene by 2.

#n = 6.11 xx 10^-3 xx 2#
#n = 1.22 xx 10^-2# moles

We can use the number of moles to calculate the volume of #CO_2# by rearranging the ideal gas law for volume:

#V = (nRT) / p#
#V = (1.22 xx 10^-2 xx 8.314 xx 317.15) / (0.989)#
#V = 32.53 L#