# The combustion of propane...?

Jul 5, 2018

You have the stoichiometry...

#### Explanation:

${C}_{3} {H}_{8} \left(g\right) + 5 {O}_{2} \left(g\right) \rightarrow 3 C {O}_{2} \left(g\right) + 4 {H}_{2} O \left(l\right) + \Delta$

$\text{Moles of propane} = \frac{2200 \cdot g}{44.1 \cdot g \cdot m o {l}^{-} 1} = 49.9 \cdot m o l$

And clearly, by the stoichiometry if $50 \cdot m o l$ propane are COMPLETELY combusted...$150 \cdot m o l$ carbon dioxide results, and $250 \cdot m o l$ dioxygen are required...

i.e. a mass of $49.9 \cdot m o l \times 5 \times 32.00 \cdot g \cdot m o {l}^{-} 1 \cong 8 \cdot k g$ ...