Answer:
587.5 L (Option A)
Explanation:
The reaction is:
2 C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O
We know that in STP conditions 1 mol of any gas is contained in 22.4L
Then, we can make a rule of three, to determine the moles of produced carbon dioxide.
22.4L is the volume for 1 mol
376 L will be the volume for (376 . 1) 22.4 = 16.78 moles at STP conditions.
Stoichiometry is 16:25.
16 moles of CO₂ are produce by the reaction of 25 moles of O₂
Then, 16.78 moles of CO₂ were produced by (16.78 . 25) /16 = 26.2 moles.
Now, the rule of three again.
1 mol of oxygen gas is contained at 22.4L, at STP conditions
26.2 moles might be contained at (26.2 . 22.4)/1 = 587.5 L
