Question

Give the theoretical yield, in moles, of CO₂ from the reaction of 4.00 moles of C8H18 with 4.00 moles of O₂. 2 C8H18 + 25 O₂ → 16 CO₂ + 18 H₂O

1) 2.56 moles
2) 64.0 moles
3) 0.640 moles
4) 16.0 moles

Answer 1

The theoretical yield of CO₂ from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂ is 2.56 moles, as oxygen is the limiting reactant.

Step-by-step explanation:

The theoretical yield, in moles, of CO₂ from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂ is determined by looking at the balanced chemical equation:

2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O(g)

According to the equation, 2 moles of octane (C₈H₁₈) react with 25 moles of oxygen (O₂) to produce 16 moles of carbon dioxide (CO₂). Since you have 4 moles of octane available, it would normally react with 50 moles of O₂ to produce 32 moles of CO₂. However, because only 4 moles of O₂ are available, oxygen is the limiting reactant. To find out how much CO₂ is produced from 4 moles of the limiting reagent O₂, you would set up a proportion:

25 moles of O₂ produce 16 moles of CO₂

4 moles of O₂ would produce X moles of CO₂

By solving this proportion, you get:

(4 moles O₂ / 25 moles O₂) × 16 moles CO₂ = (4/25) × 16 = 2.56 moles CO₂

Therefore, the theoretical yield of CO₂ when reacting 4.00 moles of O₂ with an excess of C₈H₁₈ is 2.56 moles of CO₂. This is less than the maximum possible yield of 32 moles CO₂ due to the limiting amount of oxygen.