# 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?

Nov 20, 2017

32.53 L of $C {O}_{2}$

#### Explanation:

Treating the gases involved in this reaction as ideal gases means we can solve this problem using the Ideal Gas Law $p V = n R T$
Where $p$ = pressure (atm), $V$ = volume (L), $n$ = number of moles, $T$ = temperature (K), and $R$ = the gas constant ($8.314 J {K}^{-} 1 m o {l}^{-} 1$)

First calculate the number of moles of ethylene gas we are burning by rearranging the ideal gas law for n:
$n = \frac{p V}{R T}$

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

${21}^{\circ} C$ = $294.15 K$

$n = \frac{13.8 L \times 1.083}{8.413 \times 294.15}$

$n = 6.11 \times {10}^{-} 3$ moles

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

$n = 6.11 \times {10}^{-} 3 \times 2$
$n = 1.22 \times {10}^{-} 2$ moles

We can use the number of moles to calculate the volume of $C {O}_{2}$ by rearranging the ideal gas law for volume:

$V = \frac{n R T}{p}$
$V = \frac{1.22 \times {10}^{-} 2 \times 8.314 \times 317.15}{0.989}$
$V = 32.53 L$