   # Ethane bums in air to give H 2 O and CO 2 . 2 C 2 H 6 (g) + 7 O 2 (g) → 4 CO 2 (g) + 6 H 2 O(g) (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) A 3.26-L flask contains C 2 H 6 at a pressure of 256 mm Hg and a temperature of 25 °C. Suppose O 2 gas is added to the flask until C 2 H 6 and O 2 are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of O 2 and what is the total pressure in the flask? ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 10, Problem 68GQ
Textbook Problem
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## Ethane bums in air to give H2O and CO2.2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(g) (a) Four gases are involved in this reaction. Place them in order of increasing rms speed. (Assume all are at the same temperature.) (b) A 3.26-L flask contains C2H6 at a pressure of 256 mm Hg and a temperature of 25 °C. Suppose O2 gas is added to the flask until C2H6 and O2 are in the correct stoichiometric ratio for the combustion reaction. At this point, what is the partial pressure of O2 and what is the total pressure in the flask?

(a)

Interpretation Introduction

Interpretation:

For the given combustion reaction the gases produced should be arranged in the order of their increasing rms speed has to be determined.

Concept introduction:

The root mean square velocity μ is defined as the measure of velocity of particle in gas.  It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

μrms=(3RTM)1/2 (1)

(gas constant)R=8.314JKmolM=Molarmass

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Partial pressure: The partial pressure for any gas can be obtained by multiplication of total pressure of the gas with the mole fraction of the gas present in that total mixture.

### Explanation of Solution

The expression used to calculate the rms speed is as follows,

μrms=(3RTM)1/2

Both the given gases differ only in the molar mass values since the temperature are same.

From the above equation it is clear that gas with higher molar weight will have lower velocity.

Therefore, the gas should be arranged considering their molecular weight as follows,

Molar mass of H2=18.02g/molMolar mass of C2H6=30

(b)

Interpretation Introduction

Interpretation:

For the given the partial pressure for O2 under given conditions has to be determined.

Concept introduction:

The root mean square velocity μ is defined as the measure of velocity of particle in gas.  It is the method to determine the single velocity value for particles.

Root mean square velocity can be determined,

μrms=(3RTM)1/2 (1)

(gas constant)R=8.314JKmolM=Molarmass

Molar mass: The molar mass of a substance is determined by dividing the given mass of substance by the amount of the substance.

Partial pressure: The partial pressure for any gas can be obtained by multiplication of total pressure of the gas with the mole fraction of the gas present in that total mixture.

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