Question

Give the percent yield when 28.16 g of CO2 are formed from the reaction of 8.000 moles of C8H18 with 16.00 moles of O2.

Answer 1

Percent yield of CO₂ is 6.25 %.

Step-by-step explanation:

Given data:

Percent yield of CO₂ = ?

Actual yield of CO₂ = 28.16 g

Number of moles of C₈H₁₈ = 8.000 mol

Number of moles of O₂ = 16.00 mol

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Now we will compare the moles of C₈H₁₈ and O₂ with CO₂.

C₈H₁₈ : CO₂

2 : 16

8.000 : 16/2×8.000 = 64 mol

O₂ : CO₂

25 : 16

16 : 16/25×16= 10.24 mol

Less number of moles of CO₂ are produced from 16 moles of O₂. it will limit the yield of CO₂.

Grams of CO₂ produced:

Mass = number of moles × molar mass

Mass = 10.24 mol × 44 g/mol

Mass = 450.56 g

Percentage yield of CO₂:

Percentage yield = actual yield / theoretical yield × 100

Percentage yield = 28.16 g/ 450.56 g× 100

Percentage yield = 6.25 %