At 0 degrees C, a 1.0-L flask contains 5.0×10−2 mol N2, 1.5×102 mg O2, and 5.0×1021 molecules of NH3. What is the partial pressure of each gas, and what is the total pressure in the flask?

Answered: At 0 degrees C, a 1.0-L flask contains…

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter6: The Gaseous State

Section: Chapter Questions

Problem 6.86QE

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At 0 degrees C, a 1.0-L flask contains 5.0×10⁻² mol N2, 1.5×10² mg O₂, and 5.0×10²¹ molecules of NH₃. What is the partial pressure of each gas, and what is the total pressure in the flask?

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$G i v e n, v o l u m e o f f l a s k V = 1.0 L t e m p e r a t u r e T = 0 ° C = 273.15 K n u m b e r o f m o l e s o f N_{2} = 5.0 \times 10^{- 2} m o l = 0.05 m o l m a s s o f O_{2} = 1.5 \times 10^{2} m g = 150 m g = 0.15 g r a m n u m b e r o f m o l e c u l e s o f N H_{3} = 5.0 \times 10^{21} w e a r e a s k e d t o c a l c u l a t e t h e p a r t i a l p r e s s u r e o f e a c h g a s a n d t h e t o t a l p r e s s u r e i n f l a s k . N o w, t h e n u m b e r o f m o l e s o f O_{2} a n d N H_{3} a r e, N o . o f m o l e s o f O_{2} = \frac{m a s s o f O_{2}}{m o l a r m a s s o f O_{2}} N o . o f m o l e s o f O_{2} = \frac{0.15 g r a m}{32 \frac{g}{m o l}} = 0.00468 m o l a g a i n, w e k n o w f r o m m o l e c o n c e p t, 6.023 \times 10^{23} m o l e c u l e s o f N H_{3} = 1 m o l o f N H_{3} s o, 1 m o l e c u l e s o f N H_{3} = \frac{1}{6.023 \times 10^{23}} m o l o f N H_{3} t h u s, 5.0 \times 10^{21} m o l e c u l e s o f N H_{3} = \frac{5.0 \times 10^{21} \times 1}{6.023 \times 10^{23}} m o l o f N H_{3} = 0.83015 \times 10^{21 - 23} m o l o f N H_{3} = 0.0083 m o l o f N H_{3}$

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