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 CO_2

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 (8.314 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