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(l) + 25 O2(

Answer 1

2C₈H₁₈ + 25O₂ = 16CO₂ + 18H₂O

n(C₈H₁₈)=4 mol
n(O₂)=4 mol

C₈H₁₈:O₂=2:25 ⇒ 4:50  oxygen is the limiting reagent

n(CO₂)=16n(O₂)/25

n(CO₂)=16*4/25=2.56 mol

Answer 2

Answer:

Theoretical yield = 2.56  moles of CO₂

Explanation:

2 C₈H₁₈(l) + 25 O₂( g) → 16 CO₂( g) + 18 H₈O( g)

As per the given balanced equation 2 moles of C₈H₁₈ combines with 25 moles of O₂. So 4 moles of of C₈H₁₈ requires 50 moles of O₂. But in the given question we have 4 moles of O₂ there O₂ is the limiting reagent.

The limiting reagent determines the yield of the product. Therefore, here O₂ will determines the yield of CO₂.

As per the given balanced equation 25 moles of O₂ yields 16 moles of CO₂.

Therefore, 4 moles of O₂ will yield ($\frac{16}{25}$ x 4 ) moles of CO₂

= 2.56  moles of CO₂