Calculate the heat released (kJ) in the reaction of 3.50 g of acetylene (C2H2) and excess hydrogen gas to form ethane gas: C2H2(g) + 2H2(g) –→ C2H6(g) Given: 2C2H2(g) + 502(g) → 4CO2(9) + 2H2O(g) AH = -2320 kJ/mol 2C2H6(g) + 702(g) → 4CO2(g) + 6H2O(g) AH = –3040 kJ/mol 2H2(g) + O2(g) → 2H2O(g) AH = -572 kJ/mol %3D
Calculate the heat released (kJ) in the reaction of 3.50 g of acetylene (C2H2) and excess hydrogen gas to form ethane gas: C2H2(g) + 2H2(g) –→ C2H6(g) Given: 2C2H2(g) + 502(g) → 4CO2(9) + 2H2O(g) AH = -2320 kJ/mol 2C2H6(g) + 702(g) → 4CO2(g) + 6H2O(g) AH = –3040 kJ/mol 2H2(g) + O2(g) → 2H2O(g) AH = -572 kJ/mol %3D
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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
Transcribed Image Text:29.5 kJ
None of the above
27.5 kJ
30.5 kJ
28.5 kJ

Transcribed Image Text:Calculate the heat released (kJ) in the reaction of 3.50 g of acetylene (C2H2) and excess hydrogen gas to form
ethane gas:
C2H2(g) + 2H2(g) –→ C2H6(g)
Given:
2C2H2(g) + 502(g) → 4CO2(g) + 2H2O(g) AH = -2320 kJ/mol
2C2H6(g) + 702(g) → 4CO2(g) + 6H20(g) AH = -3040 kJ/mol
2H2(g) + O2(g) → 2H20(g) AH = -572 kJ/mol
Multiple Choice
29.5 kJ
None of the above
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