Question

Air contains 20% O2 by volume. How much volume of air will be required for complete combustion of 100 cc of acetylene gas?

Answer 1

Explanation:

equation:

2C2H2+5O2⟶4CO2+2H2O

From stoichiometric calculations:

2 C2H2 moles will react with 5 molecules of oxygen

1 C2H5⟶5/2 moles of O2

22.4 l C2H2⟶ 5/2×22.4 l of O2

1 volume of air -> 0.2 (20/100) volume of oxygen

Now the only thing I don’t know is how to convert the volume into liters so that I can find the correct answer.

Answer 2

Given:

Percentage of oxygen in air = 20%

Volume of acetylene = 100 cc = 100 ml = 0.1 L

To Find:

The volume of air that will be required for the complete combustion of 100 cc of acetylene gas.

Calculation:

- The combustion of acetylene can be given as:

C2H2 + 5/2 O2 → 2 CO2 + H2O

⇒ Oxygen required for 1 mole of C2H2 = 5/2 = 2.5 mole

⇒ Oxygen required for 22.4 L of C2H2 = 2.5 × 22.4 = 56 L

⇒ Oxygen required for 0.1 L of C2H2 = (56/22.4) × 0.1 = 0.25 L

- According to question:

Percentage of oxygen in air = 20 %

⇒ The volume of air required for 0.2 L of oxygen = 1 L

⇒ The volume of air required for 0.25 L of oxygen = (1/0.2) × 0.25 = 1.25 L

- So, 1.25 L of air will be required for the complete combustion of 100 cc of acetylene gas.