What confused me in this question was:

After absorption by KOH

You don't need to know what this does in VCE Chemistry, but a few Google searches suggested KOH reacts with water and carbon dioxide. This makes sense as a combustion with a hydrocarbon will always produce H₂O and CO₂.

First, let's account for what's gone on:

For the purposes of this question, I will convert cm³ to mL as it's easier to write

• Started off with 80 mL of oxygen, and ended up with 30 mL of oxygen (50 mL was used)
• We also began with 20 mL of hydrocarbon
• Post-explosion, pre-absorption, there was 70 mL of gases (a concoction of all the reactants and products)
• The concoction consists of: 30 mL of oxygen, so there was 40 mL of CO₂ and H₂O (we're assuming here that oxygen was in excess by 30 mL, and there's no hydrocarbon left over).

Using the fact that the volume of a gas (V) is directly proportional to the amount of the gas (n) under constant T and P (which is why they said "cooling to room temperature"), then we can say that:

• n(hydrocarbon) : n(O₂) = 20 mL : 50 mL = 2 mol : 5 mol
• Using 2 mol : 5 mol, then n(CO₂) + n(H₂O) = 4 mol (because there was 40 mL)

Let the hydrocarbon have the formula: C_xH_y.

Combustion reaction: 2C_xH_y + 5O₂ → aCO₂ + bH₂O

• We need to find a and b.
• We know that: a + b = 4 (from n(CO₂) + n(H₂O) = 4 mol)
• To balance the O, we need: 10 = 2a + b

This yields: a = 6, b = -2, which does not make sense.

I'm going to try the question again, with n(CO₂) + n(H₂O) = 7 mol (70 mL of gas, not subtracting the 30 mL of oxygen) now instead, as the question may have forgot to take that into account!

1. a + b = 7
2. 10 = 2a + b

Result: a = 3, b = 4
Reaction: 2C_xH_y + 5O₂ → 3CO₂ + 4H₂O

Therefore: x = 1.5 and y = 4, which doesn't really make sense, but that means of course the reaction should have been: 2C₃H₈ + 10O₂ → 6CO₂ + 8H₂O

Answer: C₃H₈

Kinda complicated - need to work with two balances simultaneously in order to get this one out, not to mention a sloppily written question that screwed us up the first time! (Note: solving simultaneous equations is unlikely to be required by a VCE Chemistry question - only try this if you love maths!)