Question

If 4.00 moles of gasoline are burned, what volume of oxygen is needed if the pressure is 0.953 atm, and the temperature is 35.0°C?

Answer 1

1325 L

Step-by-step explanation:

For this excersise we assume gasoline as an octane.

The combustion is:

2C₈H₁₈ + 25 O₂ → 16 CO₂ + 18H₂O

2 moles of octane react to 25 moles of oxygen, according to stoichiometry.

If we have 4 moles of gasoline, we need (4 . 25)/2 = 50 moles of oxygen.

We can apply the Ideal Gases Law to solve this.

We convert T° to Absolute Value

35°C + 273 = 308K

P . V = n . R . T

V = (n . R . T) / P → (50 mol . 0.082 mol.K/L.atm . 308K) / 0.953 atm

Volume = 1325 L