In the combustion of octane, 2 C8H18 +25 O2 --> 16 CO2 + 18 H2O, if 376.0 liters of carbon dioxide formed at STP, then the volume of oxygen used was A. 587.5 L B. 240.6 L C. 839.3 L D. 299.1 L

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

what is the volume occupied by 40 grams of argon gas (ar) at standard conditions? what volume, in liters, is occupied by 1.5 x 1023 

a is the answer to this question

explanation: