Question

The theoretical air required for the complete combustion of 5 kg of ethylene gas is: [Air: 78% $N_{2}$ and 22% $O_{2}$ by mass]

• A. 107.92 kg
• B. 27.52 kg
• C. 17.14 kg
• D. 77.92 kg

Hint:

Check the basis of air composition carefully! Most textbooks use 23% $O_2$ by mass, but this question explicitly provides 22%. Always use the values provided in the problem statement for the most accurate result.

Answer 1

The Correct Option is D

Concept: Theoretical air calculation by mass requires the balanced chemical equation to find the mass of oxygen needed, which is then converted to air mass using the given oxygen mass fraction.

Step 1: Balance the combustion equation for Ethylene (\(C_{2}H_{4}\)).
\[ C_{2}H_{4} + 3O_{2} \rightarrow 2CO_{2} + 2H_{2}O \] Molar masses:
• \(C_{2}H_{4}\) = \( (2 \times 12) + (4 \times 1) = 28~g/mol \)
• \(O_{2}\) = \( 2 \times 16 = 32~g/mol \) From the stoichiometry, 28 kg of \(C_{2}H_{4}\) requires \(3 \times 32 = 96~kg\) of \(O_{2}\).

Step 2: Calculate oxygen required for 5 kg of Ethylene.
\[ \text{Mass of } O_{2} = 5 \times \frac{96}{28} = 5 \times 3.428 \approx 17.14~kg \]

Step 3: Calculate theoretical air mass.
Given air is 22% \(O_{2}\) by mass: \[ \text{Mass of air} = \frac{\text{Mass of } O_{2}}{0.22} = \frac{17.14}{0.22} \approx 77.91~kg \]