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10.0 g of gaseous ammonia and 6.50 g of oxygen gas are introduced into a previously evacuated 5.50 L vessel. If the ammonia and oxygen then react to yield NO gas and water vapor, what is the final gas pressure inside the vessel at 23?C

Respuesta :

jacob193

3.54 × 10⁵ Pa.

Explanation

Ammonia NH₃ and oxygen gas O₂ reacts at a one-to-one molar ratio to produce NO and H₂O.

NH₃ + O₂ → NO + H₂O (Balanced)

  • NH₃ has a molar mass of 17.03 g/mol. 10.0 g of NH₃ contains 0.5872 mol molecules.
  • O₂ has a molar mass of 32.00 g/mol. 6.50 g of H₂O contains 0.2031 mol molecules.
  • H₂O is the limiting reactant. All 0.2031 mol of O₂ will be consumed.
  • Only 0.2031 mol of NH₃ will be consumed. 0.5872 - 0.2031 = 0.3841 mol of NH₃ will be in excess.
  • The reaction will produce 0.2031 mol of NO and 0.2031 mol of H₂O.

The vessel will contain

  • 0.3841 mol NH₃,
  • 0 mol O₂,
  • 0.2031 mol NO, and
  • 0.2031 mol H₂O.

It will contain 0.3841 + 0.2031 + 0.2031 = 0.7903 mol gas particles by the end of the reaction.

[tex]P = n \cdot R \cdot T / V \\\phantom{P} = 0.7903 \times (8.314 \times 10^{3}) \times (273 + 23) / 5.50 \\\phantom{P} = 3.54 \times 10^5 \; \text{Pa}[/tex]

Assuming that the final mixture is an ideal gas, it will exert a pressure of 3.54 × 10⁵ Pa on the container.

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