Final answer:
Without the density of C8H8, the volume of styrene cannot be determined; however, approximately 15 liters of oxygen would be needed to produce 5.00 liters of water vapor at STP, according to the balanced chemical equation for the combustion of styrene.
Step-by-step explanation:
To calculate the volume of styrene (C8H8) and oxygen (O2) needed to produce 5.00 L of water vapor (H2O(g)) at STP, we first need the balanced chemical equation for the combustion of styrene:
C8H8(l) + 12O2(g) → 8CO2(g) + 4H2O(g)
From the balanced equation, we see that 1 mole of C8H8 produces 4 moles of H2O. Since 1 mole of gas occupies 22.4 L at STP, producing 5.00 L of H2O(g) corresponds to 5.00 L / 22.4 L/mol = 0.2232 moles of water vapor. As the molar ratio of H2O to C8H8 is 4 to 1, we would need 0.2232 moles / 4 = 0.0558 moles of C8H8. To calculate the volume of C8H8 in liters we would need its density, which is not provided. For the volume of O2, the balanced equation tells us that 12 moles of O2 are needed for each mole of C8H8, which means we'd need 12 * 0.0558 moles = 0.6696 moles of O2. At STP, this amount of O2 corresponds to 0.6696 moles * 22.4 L/mol = 14.99 L of O2.
In summary, without the density of C8H8, we cannot calculate the volume of the liquid styrene required, but using STP conditions and the balanced chemical equation, we know that approximately 15 L of oxygen is required for the combustion process that releases 5.00 L of water vapor.