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Butane gas reacts with oxygen gas to give carbon dioxide gas and water vapor (gas). If you mix butane and oxygen in the correct stoichiometric ratio, and if the total pressure of the mixture is 392 mmHg, what is the pressure (in mmHg) of water vapor after the reaction has completed (temperature and volume do not change).

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Answer:

118.776 mmHg

Explanation:

The equation of the reaction is;

C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H20(g)

Now the mole ratio  according to the balanced reaction equation is;

1 : 6.5 : 4 : 5

Hence, the total number of moles present = 1 + 6.5 + 4 + 5 = 16.5 moles

Mole fraction of water vapour = 5/16.5 = 0.303

We also know that;

Partial pressure= mole fraction * total pressure

Partial pressure of H20(g) = 0.303 *  392 mmHg = 118.776 mmHg

Eduard22sly

The pressure (in mmHg) of water vapor is 118.78 mmHg

Balanced equation for the reaction

Butane reacts with oxygen according to the following equation

2C₄H₁₀ + 13O₂ —> 8CO₂ + 10H₂O

How to determine the mole fraction of water

  • Mole of butane = 2 moles
  • Mole of oxygen = 13 moles
  • Carbon (IV) oxide = 8 moles
  • Mole of water = 10 moles
  • Total moles = 2 + 13 + 8 + 10 = 33 moles
  • Mole fraction of water =?

Mole fraction = mole / total mole

Mole fraction of water = 10 / 33

Mole fraction of water = 0.303

How to determine the partial pressure of water

  • Mole fraction of water = 0.303
  • Total pressure = 392 mmHg
  • Partial pressure of water =?

Partial pressure = mole fraction × total pressure

Partial pressure of water = 0.303 × 392

Partial pressure of water = 118.78 mmHg

Learn more about partial pressure:

https://brainly.com/question/15577259

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