# An 8.8*g mass of propane is completely combusted. What mass of carbon dioxide will result?

Mar 31, 2017

Approx. $26 \cdot g$..........................

#### Explanation:

WE have the stoichiometric equation:

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

Which tells us unequivocally that the combustion of 1 mole of propane gives 3 moles of carbon dioxide.

And then we simply access the molar quantity of propane:

$= \frac{8.8 \cdot g}{44.10 \cdot g \cdot m o {l}^{-} 1} = 0.200 \cdot m o l$

And since, per the stoichiometric equation, if $0.200 \cdot m o l$ $\text{propane}$ is completely combusted, 3xx0.200*molxx44.10*g*mol^-1=??*g $C {O}_{2} \left(g\right)$ will result.

At normal pressure and temperature, $1 \cdot a t m$, and $298 \cdot K$, what volume will the evolved gas occupy?

Mar 31, 2017

24 litres at RTP (298K) or 22.4 litres at STP (273K)

#### Explanation:

1) The balanced equation gives us the mole ratio propane:oxygen which is 1:5 - this means that every mole of propane will require 5 moles of oxygen for complete combustion.

2) However we don't have one mole of propane, we have 8.8g which, using moles =mass/molar mass, equals 8.8/44 = 0.2 moles. (RMM of propane = 12x3 + 1x8 = 44, therefore molar mass = 44g/mol).

3) Applying the mole ratio, it follows that 0.2 moles of propane will react with 0.2x5 = 1 mole of oxygen.

4) One mole of any gas (assuming ideal behaviour, which is a perfectly reasonable assumption, especially with a gas like oxygen with small and non-polar molecules) occupies 24 litres at RTP and 22.4 at STP. Presto!