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Determine the volume of oxygen gas that is required to burn 4.90 mol of pentane (C5H12) at 103 kPa and 30°C. The reaction is represented by the following balanced chemical equation: C5H12 +8 02 - 5 CO2 + 6H₂O

Determine the volume of oxygen gas that is required to burn 4.90 mol of pentane (C5H12) at 103 kPa and 30°C. The reaction is represented by the following balanced chemical equation: C5H12 +8 02 - 5 CO2 + 6H₂O

Chemistry
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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Determine the volume of oxygen gas that is required to burn 4.90 mol of pentane (C5H12) at 103 kPa and 30°C. The reaction is represented by the following balanced chemical equation: C5H12 +8 02-5 CO2 + 6 H₂O
Transcribed Image Text:Determine the volume of oxygen gas that is required to burn 4.90 mol of pentane (C5H12) at 103 kPa and 30°C. The reaction is represented by the following balanced chemical equation: C5H12 +8 02-5 CO2 + 6 H₂O
Expert Solution
Step 1: Stochiomertry and ideal gas equation

We can find the number of moles of oxygen gas required to burn 4.90 moles of pentane using stochiomertry then we have two calculate the volume of th oxygen gas using ideal gas equation.

Ideal gas equation is 

PV=nRT

P= pressure

T= temprature

V = volume

n=number of moles

R= universal gas constant 

R= 8.314J/K-mol

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