1. Write the balanced chemical equation:

The reaction between ethene (C₂H₄) and oxygen (O₂) produces carbon dioxide (CO₂) and water (H₂O):

C₂H₄ + 3O₂ → 2CO₂ + 2H₂O

2. Determine the mole ratio:

From the balanced equation, we see that 1 mole of C₂H₄ reacts with 3 moles of O₂.

3. Calculate the moles of O₂ needed:

Since we have 1.50 moles of C₂H₄, we need:

1.50 moles C₂H₄ × (3 moles O₂ / 1 mole C₂H₄) = 4.50 moles O₂

4. Apply the Ideal Gas Law:

At STP (Standard Temperature and Pressure):

* Temperature (T) = 273.15 K

* Pressure (P) = 1 atm

* Ideal gas constant (R) = 0.0821 L·atm/mol·K

The Ideal Gas Law is: PV = nRT

We can rearrange this to solve for volume (V):

V = nRT / P

5. Calculate the volume of O₂:

V = (4.50 moles)(0.0821 L·atm/mol·K)(273.15 K) / (1 atm)

V ≈ 101 L

Therefore, you need approximately 101 liters of oxygen at STP to react with 1.50 moles of ethene.