Introductory Chemistry: A Foundation
Introductory Chemistry: A Foundation
9th Edition
ISBN: 9781337399425
Author: Steven S. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Textbook Question
Chapter 13, Problem 98QAP

Dinitrogen monoxide, N 2 O , reacts with propane, C 3 H 8 , to farm nitrogen, N 2 ; carbon dioxide. CO 2 ; and water, H 2 O .

Write a balanced chemical equation for this reaction, treating all substances as gases. Include phases in your equation.

Two reservoirs, separated by a closed valve (of negligible volume), are filled with the reactants (one on each side). The pressure of N 2 O is measured to be 1 . 4 atm. The pressure of C 3 H 8 is measured to be 1 . 2 atm. Each reservoir has a capacity of 1 .0  L , and the temperature of the system is

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e valve between the reservoirs is opened, and the chemical reaction proceeds. Assuming ideal gas behavior and constant temperature, what is the partial pressure of CO 2 after the reaction is complete?

What is the partial pressure of N 2 O after the reaction is complete?

Expert Solution
Check Mark
Interpretation Introduction

(a)

Interpretation:

To write the balanced chemical equation between nitrogen monoxide and propane that results in the production of N2, CO2 and H2 O gases.

Concept Introduction:

According to Dalton’s atomic theory matter cannot be destroyed or created; in order to be consistent with this, one must make sure that the number of atoms for each distinct type of an element must be equal on both sides of a chemical reaction.

Answer to Problem 98QAP

The balanced chemical equation between nitrogen monoxide and propane that results in the production of N2, CO2 and H2 O gases is:

10 N2O(g) +  C3H8(g)10 N2(g) + 3 CO2(g)+ 4H2O(g).

Explanation of Solution

Nitrogen monoxide and propane are the two reactants while N2, CO2 and H2 O are the products. The chemical equation with their phases is:

 N2O(g) +  C3H8(g)N2(g) +  CO2(g)+ H2O(g)

The above reaction is not balanced as the number of atoms of oxygen, carbon, and hydrogen is not same on both the sides of the reaction.

So, in order to balance the reaction, the  N2O on the reactant side is multiplied by 10, and N2,   CO2and H2O are multiplied by 10, 3, and 4 respectively on the product side.

Thus, the balanced reaction is:

10 N2O(g) +  C3H8(g)10 N2(g) + 3 CO2(g)+ 4H2O(g).

Expert Solution
Check Mark
Interpretation Introduction

(b)

Interpretation:

The partial pressure of CO2 after the reaction undergoes completion should be determined.

Concept Introduction:

If there is a mixture of gasses in a container the total pressure exerted is the sum of partial pressure of the gasses present in the container. Partial pressure is defined as the pressure exerted by the gas if the gas was alone in the container.

Ptotal = P1 + P2 + P3where Ptotal =total pressurePi =partal pressure of ith gas component where i=1, 2, 3

Partial pressure can be derived as follows

PA=Ptotal × ΧAwhere PA=partial pressure of APtotal =total pressureΧA= mole fraction of A.

Answer to Problem 98QAP

0.21 atm.

Explanation of Solution

The balanced chemical equation can be written as follows,

10 N2O(g) +  C3H8(g)10 N2(g) + 3 CO2(g)+ 4H2O(g)(A)

It is clearly specified in the question that all three main products are in gas state,

If N2O(g) behaves ideally in the reservoir, applying the ideal gas law,

PV=nRT (1) where      P=Presssure     V=volume   T=Kelvin Temperature R=Universal gas constantand  n=number of moles

Before the vessel between the reservoirs were opened,  N2O(g) is confined to one reservoir.Hence;V=1.0 L   T=450 C0=(450+273)K=723KP=1.4 atm   R=0.08206 (L atmK mol)Rearranging equation(1) one can get  n = PVRT®(2)Substituting the above values to equation(2);nN2O(g) =1.4 atm x1.0 L0.0826(L atmK mol)×723 K =0.024 mol(sig.figures)Before the vessel between the reservoirs were opened,  C3H8(g) is confined to one reservoir.Hence;V=1.0 L   T=450 C0=(450+273)K=723KP=1.2 atm   R=0.08206 (L atmK mol)Rearranging equation(1) one can get  n = PVRT®(2)Substituting the above values to equation(2);nC3H8(g) =1.2 atm x1.0 L0.0826(L atmK mol)×723 K =0.020 mol(sig.figures)

Hence it is clear that the two reactants are not mixed exactly in the stoichiometric proportions.Therefore it is important to find the limiting reagent before proceeding with any calculations.According to the stoichiometry of equation (A);N2O(g) 10 molC3H8(g) 1molN2O(g) 0.024 mol?The amount of C3H8(g) that reacts with N2O(g) (0.024 mol)=C3H8(g) 1molN2O(g) 10 mol×N2O(g) 10 mol=0.0024 mol C3H8(g) Therefore it is clear that N2O(g) is the limiting reagent under this reaction conditions.The number of C3H8(g) that is left in the vessel without reacting with N2O(g) =(0.020-0.0024) mol=0.0176 molAccording to the stoichiometry of equation (A);C3H8(g) 1molN2(g) 10 mol0.0024 mol C3H8(g)?The amount of N2(g) formed during the reaction=N2(g) 10 molC3H8(g) 1mol×0.0024 mol C3H8(g)=0.024 mol N2(g)C3H8(g) 1molºCO2(g) 3 mol

0.0024 mol C3H8(g)?The amount of N2(g) formed during the reaction=CO2(g) 3 molC3H8(g) 1mol×0.0024 mol C3H8(g)=0.0072 mol CO2(g)C3H8(g) 1molºH2O (g) 4 mol0.0024 mol C3H8(g)?The amount of N2(g) formed during the reaction= H2O (g) 4 molC3H8(g) 1mol×0.0024 mol C3H8(g)=0.0096 mol H2O (g)The total number of gas molecules that remain in the reservoir after completion of the reaction= (0.0176 + 0.0096 +0.0072 +0.024) mol=0.0584 mol

Now,  assuming that all the gas species left in the total reservoirs behaves ideally using the ideal gas law equation(1)n =0.0576 mol  T=450 C0=(450+273)K=723K(temperature doesn't change in the process)R=R=0.08206 (L atmK mol) V=2.0 Lrearranging equation (1) we havePT = nRTV where PT=total pressure in the reaction vessel after reaction is completed®(3)Substituting the above values to (3)PT=0.0584 mol×0.0826(L atmK mol)×723 K2.0 L=1.73 atm(sig.figures)Let partial pressure of CO2(g) after reaction undergoes completion = PCO2(g)PCO2(g)=PT× mole fraction of CO2(g) (QDalton's law of partial pressure)PCO2(g)=PT×moles of CO2(g) total moles (g) substituting PT=1.73 atm and moles of CO2(g) =0.0072 mol total moles=0.0584 molPCO2(g)= 1.73 atm ×0.0072 mol0.0584 mol=0.21 atm.

Expert Solution
Check Mark
Interpretation Introduction

(c)

Interpretation:

The partial pressure of CO2 after the reaction undergoes completion should be calculated.

Concept Introduction:

If there is a mixture of gasses in a container the total pressure exerted is the sum of partial pressure of the gasses present in the container. Partial pressure is defined as the pressure exerted by the gas if the gas was alone in the container.

Ptotal = P1 + P2 + P3where Ptotal =total pressurePi =partal pressure of ith gas component where i=1, 2, 3

Partial pressure can be derived as follows

PA=Ptotal × ΧAwhere PA=partial pressure of APtotal =total pressureΧA= mole fraction of A.

Answer to Problem 98QAP

Zero.

Explanation of Solution

Let partial pressure of N2O(g) after reaction undergoes completion = PN2O(g)

PN2O(g)=PT× mole fraction of N2O(g) (QDalton's law of partial pressure)PN2O(g)=PT×moles of N2O(g) total moles (g) 

N2O(g) is the limiting agent.

Hence it is clear that there is no N2O(g) after the reaction undergoes the completion.

moles of N2O(g) =0.0000 mol

substituting PT=1.73 atm and moles of N2O(g) =0.0000 mol total moles=0.0584 molPCO2(g)= 1.73 atm ×0.0000 mol0.0584 mol=0 atm.

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Chapter 13 Solutions

Introductory Chemistry: A Foundation

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