2. The hydrocarbon benzene combusts to give CO2 and water vapor, and the unbalanced equation is: C6H6(2) + O2l8) → CO2(8) + H2O(g) If 0.095 g benzene burns completely in O2, what is the pressure of water vapor in a 4.75 L flask at 30.0°C?

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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Tip for handling stoichiometry problems involving gases: Use the ideal gas law (PV = nRT) to solve for the number of moles (n) of gas involved. Once you have a value of moles, you can then use mole ratios and mole conversion calculations, just as you would with any other stoichiometry problem! This also works the other way around; with a number of moles of gas, you can solve for the gas pressure (P), volume (V), or temperature in Kelvin (T) with the ideal gas law, as specified by the problem.

2.
The hydrocarbon benzene combusts to give CO2 and water vapor, and the unbalanced
equation is:
C6H6(g) +
O2(g)
CO2{g) +
H2O(g)
If 0.095 g benzene burns completely in O2, what is the pressure of water vapor in a 4.75
L flask at 30.0°C?
Transcribed Image Text:2. The hydrocarbon benzene combusts to give CO2 and water vapor, and the unbalanced equation is: C6H6(g) + O2(g) CO2{g) + H2O(g) If 0.095 g benzene burns completely in O2, what is the pressure of water vapor in a 4.75 L flask at 30.0°C?
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