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?

1 Answer
Nov 20, 2017

32.53 L of CO2

Explanation:

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 (8.314JK1mol1)

First calculate the number of moles of ethylene gas we are burning by rearranging the ideal gas law for n:
n=pVRT

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

21C = 294.15K

n=13.8L×1.0838.413×294.15

n=6.11×103 moles

Looking at the balanced equation, ethylene gas is in a 1 : 2 molar ratio with CO2 so we can calculate the number of moles of CO2 produced when ethylene burns completely by multiplying the no. moles of ethylene by 2.

n=6.11×103×2
n=1.22×102 moles

We can use the number of moles to calculate the volume of CO2 by rearranging the ideal gas law for volume:

V=nRTp
V=1.22×102×8.314×317.150.989
V=32.53L