Show the calculation of the volume of H₂O gas formed by the combustion of 18.6 grams of C6H6 at 30°C and 1.10 atm? The combustion of benzene (C6H₁) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) 12 CO2 (g) + 6 H₂O (g)
Show the calculation of the volume of H₂O gas formed by the combustion of 18.6 grams of C6H6 at 30°C and 1.10 atm? The combustion of benzene (C6H₁) takes place by the following reaction equation. 2 C6H6 (g) + 15 O2 (g) 12 CO2 (g) + 6 H₂O (g)
Chemistry
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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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Show the calculation of the volume of H₂O gas formed by the combustion
of 18.6 grams of C6H6 at 30°C and 1.10 atm? The combustion of
benzene (CH) takes place by the following reaction equation.
2 C6H₁ (9) + 15 0₂ (9)
12 CO2 (g) + 6
H₂O (g)
Your Answer:
V = 5.377 Liters rounded to 5.38 Liters
V = nRT/P
n = .2378 moles
R= constant @ 0.0821
T= 30°C +273 = 303 K
P= 1.10 atm
V = (.2378* 0.0821 * 303) / 1.10 atm = 5.377 Liters rounded to 5.38
Liters
Moles (n) of C6H6 = g/mW --> 18.6 / 78.108 = 2378 moles C6H₁
MW (6 * 12.01) + (6 * 1.008) = 78.108 MW
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