How many liters of oxygen are needed to react with 2.878 g of octane at STP?
2C₈H₁₈(g) + 25O₂(g) → 16CO₂(g) +18H₂O(g)

Question

How many liters of oxygen are needed to react with 2.878 g of octane at STP? 2C₈H₁₈(g)+ 25O₂(g)→ 16CO₂(g)+18H₂O(g) A) 8.21 L B) 7.04 L C) 6.72 L D) 7.68 L

Quizplus · Accepted Answer

From the balanced equation, we can see that for every 25 moles of oxygen, 2 moles of octane react. Therefore, we need to find the number of moles of octane present in 2.878 g.

moles of octane = mass/molecular weight = 2.878/114.23 = 0.0252 mol

Using the mole ratio from the balanced equation, we can find the number of moles of oxygen needed:

moles of oxygen = (25/2) x moles of octane = (25/2) x 0.0252 = 0.315 mol

At STP, 1 mole of gas occupies 22.4 L. Therefore, the volume of oxygen needed is:

volume of oxygen = moles x 22.4 L/mol = 0.315 x 22.4 = 7.04 L

Therefore, the answer is B.

How Many Liters of Oxygen Are Needed to React with 2.878