Explanation To determine the number of moles of CH 4 , use ideal gas law . Before determining the number of moles, convert the given pressure 732 Torr into atmospheres by the use of following relation: ∵ 1 Torr = 1 mmHg As 760 mmHg = 1 atm, 760 Torr = 1 atm ∴ 732 Torr = 1 atm × 732 Torr 760 Torr = 0.963 atm Use ideal gas law to calculate the moles of CH 4 : P V = n R T ⇒ n = P V R T Substitute P = 0 .963 atm , V CH 4 = 25.5 L , R = 0.08206 L . atm/mol/K , and T = 298.15 K in the above expression to get n = 0 .963 atm × 25 .5 L 0 .08206 L . atm mol . K × 298 .15 K = 1 .00369 mol Similarly, determine the number of moles of H 2 O ( g ) by the use of ideal gas law. Before determining the number of moles, convert the given pressure 702 T o r r into atmospheres by the use of following relation: ∵ 1 Torr = 1 mmHg As 760 mmHg = 1 atm, 760 Torr = 1 atm ∴ 702 Torr = 1 atm × 702 Torr 760 Torr = 0.92368 atm Use ideal gas law to calculate the moles of H 2 O ( g ) : P V = n R T ⇒ n = P V R T Substitute P = 0 .92368 atm , V H 2 O ( g ) = 22.8 L , R = 0.08206 L . atm/mol/K , and T = 398.15 K in the above expression to get n = 0 .92368 atm × 22 .8 L 0 .08206 L . atm mol . K × 398 .15 K = 0 .64458 mol Determine the limiting reactant in order to calculate the volume of H 2 ( g ) produced during the reaction. The given reaction is C H 4 ( g ) + H 2 O ( g ) → C O ( g ) + 3 H 2 ( g ) As per the reaction 1 mol CH 4 reacts with 1 mol H 2 O ∴ 1.0039 mol CH 4 reacts with = 1 mol H 2 O × 1

Chemistry: A Molecular Approach Plus Mastering Chemistry with Pearson eText -- Access Card Package (4th Edition) (New Chemistry Titles from Niva Tro)

4th Edition

ISBN: 9780134103976

Author: Nivaldo J. Tro

Publisher: PEARSON

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Ch. 5 - Prob. 1SAQ Ch. 5 - Prob. 2SAQ Ch. 5 - Prob. 3SAQ Ch. 5 - Prob. 4SAQ Ch. 5 - Prob. 5SAQ Ch. 5 - Prob. 6SAQ Ch. 5 - Prob. 7SAQ Ch. 5 - Prob. 8SAQ Ch. 5 - Q9. A gas sample at STP contains 1.15 g oxygen and...Ch. 5 - Prob. 10SAQ

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