Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 6, Problem 6.175AP
Interpretation Introduction

Interpretation: Wetlands are given to be used to treat agricultural waste from rain runoff and groundwater by the technique denitrification which converts nitrate ions to nitrogen gas. The volume of N2 produced at 17°C and 1atm if the denitrification process is complete and the volume of CO2 produce is to be calculated. If the total pressure is assumed to be 1atm then the density of the mixture is to be calculated.

Concept introduction: The Charles law states that “the volume of a given mass of gas varies directly with the absolute temperature of the gas when pressure is kept constant”.

To determine: The volume of N2andCO2 produced and the density of the mixture if the PTotal of the system is assumed to be 1atm .

Expert Solution & Answer
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Answer to Problem 6.175AP

Solution

The volume of N2andCO2 produced is 38.38L_ and 191.84L_ respectively. The density of mixture is 1.73g/L_ .

Explanation of Solution

The given equation is,

2NO3(g)+5CO(g)+2H+(aq)N2(g)+H2O(l)+5CO2(g)

Here, 200 g of NO3 are reacting.

The number of moles of NO3 is calculated by the formula,

n=MassingramsMolarmassingrams

Substitute the given values of mass and molar mass in the above expression.

n=20062=3.225mol

Molar mass of NO3 is 3.225g/mol .

Since the molecular mass 2×3.225g/mol of NO3 produces 1molofN2 .

Therefore 2mol of NO3 produces 1 mole of N2 =1×3.2252=1.6125molofN2

Thus, the amount of N2 produced will be 1.6125mol .

The one mole of any gas occupies 22.4L at STP.

So, the volume occupied by 1.6125molofN2 at STP =22.41×1.6125=36.12L

The volume of the N2 gas is calculate by the formula,

V1T1=V2T2

Where,

  • T1 is the initial temperature 290K .
  • T2 is the new final temperature 17°C .
  • V1 is the initial volume 36.12L .
  • V2 is the new final volume.

The conversion of temperature from °CtoK is done as,

K=°C+273.15

For T2

T2=17°C+273.15=290.15K290K

Substitute these values of T1 , T2 and V1 in above expression.

V1T1=V2T2V2=36.12L×290K273K=38.38L_

Therefore, the volume of N2 at 17°C is 38.38L_ .

2molofNO3produces5molofCO2gas.

Molar mass of NO3 is 3.225g/mol .

Since the molecular mass 2×3.225g/mol of NO3 produces 5molofCO2 .

Therefore 2mol of NO3 produces 5 mole of CO2 =5×3.2252=8.0625molofCO2

Thus, the amount of CO2 produced will be 8.0625mol .

The one mole of any gas occupies 22.4L at STP.

So, the volume occupied by 8.0625molofCO2 at STP =22.41×8.0625=180.6L

The volume of the CO2 gas is calculated by the formula,

V1T1=V2T2

Where,

  • T1 is the initial temperature 290K .
  • T2 is the new final temperature 17°C .
  • V1 is the initial volume 180.6L .
  • V2 is the new final volume.

The conversion of temperature from °CtoK is done as,

K=°C+273.15

For T2

T2=17°C+273.15=290.15K290K

Substitute these values of T1 , T2 and V1 in above expression.

V1T1=V2T2V2=180.6L×290K273K=191.84L_

Therefore, the volume of CO2 at 17°C is 191.84L_ .

The density of CO2 at STP is 1.96g/L and the density of N2 at STP is 1.255g/L .

The density of N2 at 17°C is calculated by the formula,

V1V2=T1T2

Where,

T1 is the initial temperature.

T2 is the new final temperature.

V1 is the initial volume.

V2 is the new final volume.

Simplify the above expression.

V1M1V2M2=T1T2 (1)

The density is given as,

MV=D

Where,

  • M is the molar mass.
  • V is the volume.
  • D is the density.

Substitute the above relation of density in equation (1).

D1D2=T1T2D1=D2T1T2

Now, substitute the given values in the above expression,

D1=D2T1T2

Substitute the values of D2 , T1 and T2 in the above expression.

D1=1.255g/L×273K290K=1.181g/L

Similarly, the density of CO2 at 17°C is calculated by the formula,

V1V2=T1T2

Where,

  • T1 is the initial temperature.
  • T2 is the new final temperature.
  • V1 is the initial volume.
  • V2 is the new final volume.

Simplify the above expression

V1M1V2M2=T1T2

The density is given as,

MV=D

Where,

  • M is the molar mass.
  • V is the volume.
  • D is the density.

Substitute the above relation of density in equation (1).

D1D2=T1T2D1=D2T1T2

Now, substitute the given values in the above expression,

D1=D2T1T2

Substitute the values of D2 , T1 and T2 in the above expression.

D1=1.96g/L×273K290K=1.849g/L

As gas mixture contains five parts of CO2 and one part of nitrogen. So the density of mixture is calculated by the formula,

Densityofmixture=5×DensityofCO2+1×DensityofN26

Substitute the values of densities of CO2 and N2 in the above expression.

Densityofmixture=5×1.84+1×1.1816=1.73g/L_

Therefore, density of mixture 1.73g/L_ .

Conclusion:

The volume of N2andCO2 produced is 38.38L_ and 191.84L_ respectively. The density of mixture is 1.73g/L_

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Chapter 6 Solutions

Chemistry

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