Question

Ethene is burnt with 150% of excess air, what is the percentage of O2 in the products?

(a) 12.5%

(b) 33.3%

(c) 45.4%

(d) 52.9%

I have been asked this question in examination.

My doubt stems from Material Balances Involving Combustion topic in chapter The Chemical Reaction Equation & Stoichometry, Material Balances for Processes Involving Reaction and Multiple Units of Basic Chemical Engineering

Answer 1

To solve this, let's first write the combustion reaction for ethene (C₂H₄) with oxygen (O₂):

Balanced Combustion Reaction:

C2H4+3O2→2CO2+2H2O\text{C}_2\text{H}_4 + 3\text{O}_2 \rightarrow 2\text{CO}_2 + 2\text{H}_2\text{O}C2​H4​+3O2​→2CO2​+2H2​O

This shows that 1 mole of ethene reacts with 3 moles of oxygen to produce carbon dioxide (CO₂) and water (H₂O).

Step 1: Determine the Oxygen Requirement

For complete combustion of 1 mole of ethene, the oxygen required is 3 moles.

Step 2: Excess Air Calculation

With 150% excess air, the total amount of air supplied is:

• The stoichiometric amount of oxygen required: 3 moles of O₂.
• Excess air is 150%, so the total amount of oxygen supplied is: Total O2=3×(1+1.5)=3×2.5=7.5 moles of O2.\text{Total O}_2 = 3 \times (1 + 1.5) = 3 \times 2.5 = 7.5 \text{ moles of O}_2.Total O2​=3×(1+1.5)=3×2.5=7.5 moles of O2​.

This means there are 7.5 moles of oxygen supplied for 1 mole of ethene.

Step 3: Oxygen in the Products

In the products, oxygen is present as:

• The oxygen that reacted with ethene (3 moles).
• The oxygen that is in excess, which remains unreacted (7.5 - 3 = 4.5 moles of O₂).

Step 4: Total Moles of Products

The products are:

• 2 moles of CO₂ (from 1 mole of ethene).
• 2 moles of H₂O (from 1 mole of ethene).
• 4.5 moles of unreacted O₂.

Thus, the total moles of products = 2 (CO₂) + 2 (H₂O) + 4.5 (O₂) = 8.5 moles.

Step 5: Percentage of O₂ in the Products

Now, we calculate the percentage of oxygen in the products:

Percentage of O2=(Moles of O2Total moles of products)×100=(4.58.5)×100=52.9%.\text{Percentage of O}_2 = \left( \frac{\text{Moles of O}_2}{\text{Total moles of products}} \right) \times 100 = \left( \frac{4.5}{8.5} \right) \times 100 = 52.9\%.Percentage of O2​=(Total moles of productsMoles of O2​​)×100=(8.54.5​)×100=52.9%.

Final Answer:

The percentage of O₂ in the products is:

(d) 52.9%.