What is the mass (in grams) of 19.00 L of propane vapor (C₃H₈) at STP? R = 0.08314 L･bar/mol･K. ChromeFileEditViewHistory Bookmarks People TabWindowHelp33%Tue 6:10 PMChem 101 HW101 Chem101+app.101edu.coEQuestion 11 of 25SubmitWhat is the mass (in grams) of 19.00 L of propane vapor (C3H8) atSTP? R= 0.08314 L·bar/mol· K.136.C789+/-х 100+PAGESW4+

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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

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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What is the mass (in grams) of 19.00 L of propane vapor (C₃H₈) at STP? R = 0.08314 L･bar/mol･K.

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Step 1:- Calculation of number of moles of propane using ideal gas equation

$S o l u t i o n : - T h e v a l u e s p r o v i d e d i n t h e q u e s t i o n a r e : - i) V o l u m e o f p r o p a n e v a p o u r, V = 19.00 L i i) u n i v e r s a l g a s c o n s \tan t, R = 0.08314 L . b a r / m o l . k i i i) M o l a r m a s s o f C_{3} H_{8} = (3 \times 12.01 + 8 \times 1.008) g / m o l = (36.03 + 8.064) g / m o l = 44.094 g / m o l A t s \tan d a r d t e m p e r a t u r e a n d p r e s s u r e (i . e . S T P), t h e t e m p e r a t u r e i s 273.15 K (o ° C, 32 ° F) a n d p r e s s u r e i s 1.013 b a r (100 K P a, 1 a t m) A c c o r d i n g t o I d e a l g a s e q u a t i o n, V = N R T; w h e r e n i s t h e n u m b e r o f m o l e s p r n = \frac{P V}{R T} i . e . n = \frac{1.013 b a r \times 19.001 L}{0.08314 L b a r m o l^{- 1} K^{- 1} \times 273.15 k} n = \frac{19.25}{22.71 m o l^{- 1}} = 0.8476 m o l [∴ \frac{1}{m o l^{- 1}} = m o l] T h u s, t h e n u m b e r o f m o l e s o f C_{3} H_{8} i s 0.8476 m o l$

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