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?

1325 L

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

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explanation: