Answer:
13.44dm^3
Explanation:
To calculate this we first need to know the number of mole produced. We will first need to balance the equation to know the theoretical mole ratio.
C2H6 (g) + 3.5O2 (g) → 2CO2 (g) + 3H2O (g)
From the balanced equation, we can deduce that one mole of ethane yielded 2 moles of carbon iv oxide. We use this information to calculate the actual number of moles yielded.
24g were reacted. Now to know the number of moles reacted, we simply divide the mass by the molar mass. The molar mass of ethane is 2(12) +6(1)= 40g/mol
The number of moles is thus 24/40 = 0.6 moles
Like we said earlier, one mole yielded 2 moles of carbon iv oxide, hence, 0.6 moles will yield 0.6 * 2 = 0.12 moles of carbon iv oxide.
Now, at stp, one mole of a gas occupies a volume of 22.4dm^3 thus, 0.6 mole will occupy 0.6 * 22.4 = 13.44dm^3
