Question

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

Answer 1

Approx. `26*g`..........................

Explanation:

WE have the stoichiometric equation:

`C_3H_8(g) + 5O_2(g) rarr 3CO_2(g) + 4H_2O(l)`

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:

`=(8.8*g)/(44.10*g*mol^-1)=0.200*mol`

And since, per the stoichiometric equation, if `0.200*mol` `"propane"` is completely combusted, `3xx0.200*molxx44.10*g*mol^-1=??*g` `CO_2(g)` will result.

At normal pressure and temperature, `1*atm`, and `298*K`, what volume will the evolved gas occupy?

Answer 2

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!