Question

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?

Answer 1

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`