   # A steel cylinder contains 5.00 moles of graphite (pure carbon) and 5.00 moles of O 2 . The mixture is ignited and all the graphite reacts. Combustion produces a mixture of CO gas and CO 2 gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0%. Calculate the mole fractions of CO, CO 2 , and O 2 in the final gaseous mixture. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 156CP
Textbook Problem
237 views

## A steel cylinder contains 5.00 moles of graphite (pure carbon) and 5.00 moles of O2. The mixture is ignited and all the graphite reacts. Combustion produces a mixture of CO gas and CO2 gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by 17.0%. Calculate the mole fractions of CO, CO2, and O2 in the final gaseous mixture.

Interpretation Introduction

Interpretation: The mole fractions of CO , CO2 and O2 in the final mixture of given reaction mixture containing 5mol graphite and 5mol O2 should be determined if the pressure of cylinder is increased by 17% after reaction at constant temperature.

Concept introduction:

• Balanced equation of a reaction is written according to law of conservation of mass.
• According toideal gas law: the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

Pn=RTV , P and n are proportional.

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.
• Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,
• moleculefractionofA,(χA=numbersof moles ofmolecule A(nA)total number of moles(nA+nB+nC

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

To determine: The mole fractions of CO , CO2 and O2 in the final mixture of given reaction mixture containing 5mol graphite and 5mol O2 .

### Explanation of Solution

Explanation

To find: the balanced equations for given reactions.

C+12O2CO

CO+O2CO2

The reactants presented are graphite C and O2 .

The products of the given reactions are COandCO2 .

Therefore,

The chemical equations for the given reactions are,

C+12O2CO

CO+O2CO2

The above chemical equations are balanced. So they are balanced chemical equations.

To find: the total number of moles of products in the given reaction.

The number of moles of products is 5.85mol .

The products of the given reactions are CO , CO2 and O2 .

Therefore, final pressure of the given reactions is depends upon the pressures of CO , CO2 and O2

The initial number of moles of given reaction is given as 5mol (moles of O2 is taken since the pressure is depended upon O2 only).

The final pressure of the given reactions is increased by 17%.

According to ideal gas law, the pressure of gas is directly proportional to the number of moles of gas at constant volume and temperature.

Pn=RTV , P and n are proportional.

Therefore,

If the final pressure of the given reactions is increased by 17%, then the final number of moles of reactions also increases by 17%.

Therefore,

The final number of moles of reactions is,

5mol×5×17100mol=5.85mol

To find: the number of moles of produced CO , CO2 and O2 in the given reaction.

The number of moles of produced CO in the given reaction is 1.70mol .

The number of moles of produced CO2 in the given reaction is 3.30mol .

The number of moles of produced O2 in the given reaction is 0.85mol .

The total number of moles of products CO , CO2 and O2 is calculated as 5.85mol .

That means,

nCO+nCO2+nO2=5.85mol -Equation_1

The number of moles of graphite is given as 5mol .

Therefore, the total number of moles of carbon element containing products is 5mol .

That means,

nCO+nCO2=5mol -Equation_2

Subtract equation 1 by 2,

That means,

nO2=0.85mol

Hence,

The number of moles of produced O2 in the given reaction is 0.85mol .

The balance equation of given reactions are

C+12O2CO

CO+O2CO2

COandO2 in the given reaction is in 1:12 ratio respectively

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