# Home barbeques are fueled with propane gas (C_3H_8). What mass of carbon dioxide (in kg) forms upon the complete combustion of 18.9 propane (approximate contents of one 5-gallon tank)?

## Assume that density of the liquid propane in the tank is 0.621 g/mL

Aug 11, 2018

Well, you have not given us all the necessary info....

#### Explanation:

${H}_{3} C - C {H}_{2} C {H}_{3} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right) + \Delta$

And, assuming you are a septic, $\text{1 US Gallon} \equiv 3.785 \cdot L$..

We combust a mass of ethane...

${\text{5 US" cancel("Gallon")xx3.785*cancelL*cancel"gallon}}^{-} 1 \times {10}^{3} \cdot \cancel{m L} \cdot \cancel{{L}^{-} 1} \times 0.621 \cdot g \cdot \cancel{m {L}^{-} 1} \equiv 11752.4 \cdot g$...

...the which represents a molar quantity with respect to ethane of...

$\frac{11752.4 \cdot g}{30.07 \cdot g \cdot m o {l}^{-} 1} = 390.8 \cdot m o l$...

And given the combustion, we get THREE equiv of carbon dioxide evolved...

$390.8 \cdot m o l \times 3 \times 44.01 \cdot g \cdot m o {l}^{-} 1 \times {10}^{3} \cdot g \cdot k {g}^{-} 1 = 51.6 \cdot k g$.

Take that atmosphere...