Ammonium nitrate can be decomposed at elevated temperatures into nitrous oxide and water vapor according to the following balanced equation. NH4NO3(s) → N2O(g) + 2 H2O(g) When 10.0 g of NH4NO3 are used in this reaction, what volume of water vapor is produced at a temperature of 255oC and a pressure of 775 mmHg?

The volume of a gas at a particular temperature and pressure can be found using ideal gas equation which is given as: PV = nRT where P is pressure V is volume n is number of moles R is Universal Gas Constant = 8.314 J/mol.K or 0.0821 atm.L/mol.K T is temperature Determine number of moles of H2O produced: Moles of NH4NO3 = Mass of NH4NO3Molar mass of NH4NO3= 10.0 g80.043 g/mol=0.125 moles ∴ As per stoichiometry of reaction, moles of H2O = 2 (Moles of NH4NO3) i.e. moles of H2O = 2×0.125 moles= 0.25 molesDetermine Volume of H2O: Pressure, P = 775 mmHg= 775 mmHg 1 atm760 mmHg= 1.02 atm Temperature, T = 255∘C =(255 + 273.15)K = 528.15 K Put these values in ideal gas equation. we get=PV = nRT(1.02 atm)×V = (0.25 moles)×(0.0821 atm.L/mol.K)×(528.15K)V = 10.63 L Hence the volume of water vapor produced is 10.63 L

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Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

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Ammonium nitrate can be decomposed at elevated temperatures into nitrous oxide and water

vapor according to the following balanced equation.

NH₄NO_3(s) → N₂O_(g) + 2 H₂O_(g)

When 10.0 g of NH₄NO₃ are used in this reaction, what volume of water vapor is produced at a

temperature of 255^oC and a pressure of 775 mmHg?

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