Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 6, Problem 6.176AP

(a)

Interpretation Introduction

Interpretation: Ammonium nitrate given to decompose on heating and the product is stated to being dependent on the reaction temperature. A sample of NH4NO3 decomposes at an unspecified temperature, and the resulting gases are collected over water at 20°C . The given questions on the basis of above statement are to be answered.

Concept introduction: 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 explanation about the statement that the volume of gases collected can be used to distinguish between the two reactions pathways.

(a)

Expert Solution
Check Mark

Answer to Problem 6.176AP

Solution

The given compound NH4NO3 decomposes to give a mixture of N2 and O2 gas, then the volume of the collected gases will be more than the volume of the other gases formed in second pathway if the reaction proceeds to create N2O .

Explanation of Solution

The given equation for NH4NO3 decomposing to give N2O is,

NH4NO3(s)N2O(g)+2H2O(g)

In this path, 1mol of N2O gas is formed and 1mol of NH4NO3 gas is formed.

This reaction proceeds to form N2 and O2 gas,

NH4NO3(s)N2(g)+12O2(g)+2H2O(g)

In this path, 1mol of N2 gas is formed and 12 mol of O2 gas is formed.

At constant temperature and pressure, the volume of gas is proportional to the numbers of moles.

Therefore, if NH4NO3 decomposes to give us a mixture of N2 and O2 gas, than the volume of the collected gases will be more than the volume of the other gases formed in second pathway if the reaction proceeds to create N2O .

(b)

Interpretation Introduction

To determine: The gas produced during the thermal decomposition of 0.256g NH4NO3 displaces 79mL of water at 20°C and 760mmHg of atmospheric pressure.

(b)

Expert Solution
Check Mark

Answer to Problem 6.176AP

Solution

The volume of gas is 79mL , it must be N2O .

Explanation of Solution

The number of moles of NH4NO3 is calculated by the formula,

n=MassingramsMolarmassingrams

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

n=0.256g80g/mol=0.0032mol

The stated reaction proceeds to 1mol of N2O gas.

The number of moles of N2O is calculated as,

1mol of NH4NO3 produces 1 mole of N2O .

Therefore, 0.0032mol NH4NO3 produces moles of N2O =(0.00321×1)mol=0.0032mol

One mole of any gas occupies 22.4L at STP.

So, the volume occupied by 0.0032molofN2O at STP =22.41×0.0032=0.07168L

The conversion of temperature from °C to K is done as,

K=°C+273.15

For T2 ,

T2=20°C+273.15=290.15K290K

The volume is calculated by the formula,

V1T1=V2T2

Where,

  • T1 is the initial temperature 273K .
  • T2 is the new final temperature 290K .
  • V1 is the initial volume 0.07168L .
  • V2 is the new final volume.

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

V1T1=V2T2V2=0.07168L×290K273K=0.077L=77mL

If the reaction proceeds to create a mixture of N2 and O2 , then 1mol of NH4NO3 will produce a mixture of O2 and N2 .

Therefore, the moles of the mixture produced from 0.0032mol of NH4NO3 is calculated as,

Molesofmixture= 0.0032molNH4NO3×32molesofmixture1molofNH4NO3=0.0048molesofmixture

One mole of any gas occupies 22.4L at STP.

So, the volume occupied by 0.0048molofmixture at STP =22.41×0.0048=0.1075L

The volume occupied by a mixture is calculated by the formula,

V1T1=V2T2

Where,

  • T1 is the initial temperature 273K .
  • T2 is the new final temperature 290K .
  • V1 is the initial volume 0.1075L .
  • V2 is the new final volume.

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

V1T1=V2T2V2=0.1075L×290K273K=0.01153L=115.3mL

As the given volume of gas is 79mL , it must be N2O .

Conclusion:

  1. a) The given compound NH4NO3 decomposes to give a mixture of N2 and O2 gas, than the volume of the collected gases will be more than the volume of the other gases formed in second pathway if the reaction proceeds to create N2O .
  2. b) The gas present is N2O

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Chapter 6 Solutions

Chemistry

Ch. 6.6 - Prob. 11PECh. 6.7 - Prob. 12PECh. 6.7 - Prob. 13PECh. 6.7 - Prob. 14PECh. 6.8 - Prob. 15PECh. 6.8 - Prob. 16PECh. 6 - Prob. 6.1VPCh. 6 - Prob. 6.2VPCh. 6 - Prob. 6.3VPCh. 6 - Prob. 6.4VPCh. 6 - Prob. 6.5VPCh. 6 - Prob. 6.6VPCh. 6 - Prob. 6.7VPCh. 6 - Prob. 6.8VPCh. 6 - Prob. 6.9VPCh. 6 - Prob. 6.10VPCh. 6 - Prob. 6.11VPCh. 6 - Prob. 6.12VPCh. 6 - Prob. 6.13VPCh. 6 - Prob. 6.14VPCh. 6 - Prob. 6.15VPCh. 6 - Prob. 6.16VPCh. 6 - Prob. 6.17VPCh. 6 - Prob. 6.18VPCh. 6 - Prob. 6.19VPCh. 6 - Prob. 6.20VPCh. 6 - Prob. 6.21VPCh. 6 - Prob. 6.22VPCh. 6 - Prob. 6.23QPCh. 6 - Prob. 6.24QPCh. 6 - Prob. 6.25QPCh. 6 - Prob. 6.26QPCh. 6 - Prob. 6.27QPCh. 6 - Prob. 6.28QPCh. 6 - Prob. 6.29QPCh. 6 - Prob. 6.30QPCh. 6 - Prob. 6.31QPCh. 6 - Prob. 6.32QPCh. 6 - Prob. 6.33QPCh. 6 - Prob. 6.34QPCh. 6 - Prob. 6.35QPCh. 6 - Prob. 6.36QPCh. 6 - Prob. 6.37QPCh. 6 - Prob. 6.38QPCh. 6 - Prob. 6.39QPCh. 6 - Prob. 6.40QPCh. 6 - Prob. 6.41QPCh. 6 - Prob. 6.42QPCh. 6 - Prob. 6.43QPCh. 6 - Prob. 6.44QPCh. 6 - Prob. 6.45QPCh. 6 - Prob. 6.46QPCh. 6 - Prob. 6.47QPCh. 6 - Prob. 6.48QPCh. 6 - Prob. 6.49QPCh. 6 - Prob. 6.50QPCh. 6 - Prob. 6.51QPCh. 6 - Prob. 6.52QPCh. 6 - Prob. 6.53QPCh. 6 - Prob. 6.54QPCh. 6 - Prob. 6.55QPCh. 6 - Prob. 6.56QPCh. 6 - Prob. 6.57QPCh. 6 - Prob. 6.58QPCh. 6 - Prob. 6.59QPCh. 6 - Prob. 6.60QPCh. 6 - Prob. 6.61QPCh. 6 - Prob. 6.62QPCh. 6 - Prob. 6.63QPCh. 6 - Prob. 6.64QPCh. 6 - Prob. 6.65QPCh. 6 - Prob. 6.66QPCh. 6 - Prob. 6.67QPCh. 6 - Prob. 6.68QPCh. 6 - Prob. 6.69QPCh. 6 - Prob. 6.70QPCh. 6 - Prob. 6.71QPCh. 6 - Prob. 6.72QPCh. 6 - Prob. 6.73QPCh. 6 - Prob. 6.74QPCh. 6 - Prob. 6.75QPCh. 6 - Prob. 6.76QPCh. 6 - Prob. 6.77QPCh. 6 - Prob. 6.78QPCh. 6 - Prob. 6.79QPCh. 6 - Prob. 6.80QPCh. 6 - Prob. 6.81QPCh. 6 - Prob. 6.82QPCh. 6 - Prob. 6.83QPCh. 6 - Prob. 6.84QPCh. 6 - Prob. 6.85QPCh. 6 - Prob. 6.86QPCh. 6 - Prob. 6.87QPCh. 6 - Prob. 6.88QPCh. 6 - Prob. 6.89QPCh. 6 - Prob. 6.90QPCh. 6 - Prob. 6.91QPCh. 6 - Prob. 6.92QPCh. 6 - Prob. 6.93QPCh. 6 - Prob. 6.94QPCh. 6 - Prob. 6.95QPCh. 6 - Prob. 6.96QPCh. 6 - Prob. 6.97QPCh. 6 - Prob. 6.98QPCh. 6 - Prob. 6.99QPCh. 6 - Prob. 6.100QPCh. 6 - Prob. 6.101QPCh. 6 - Prob. 6.102QPCh. 6 - Prob. 6.103QPCh. 6 - Prob. 6.104QPCh. 6 - Prob. 6.105QPCh. 6 - Prob. 6.106QPCh. 6 - Prob. 6.107QPCh. 6 - Prob. 6.108QPCh. 6 - Prob. 6.109QPCh. 6 - Prob. 6.110QPCh. 6 - Prob. 6.111QPCh. 6 - Prob. 6.112QPCh. 6 - Prob. 6.113QPCh. 6 - Prob. 6.114QPCh. 6 - Prob. 6.115QPCh. 6 - Prob. 6.116QPCh. 6 - Prob. 6.117QPCh. 6 - Prob. 6.118QPCh. 6 - Prob. 6.119QPCh. 6 - Prob. 6.120QPCh. 6 - Prob. 6.121QPCh. 6 - Prob. 6.122QPCh. 6 - Prob. 6.123QPCh. 6 - Prob. 6.124QPCh. 6 - Prob. 6.125QPCh. 6 - Prob. 6.126QPCh. 6 - Prob. 6.127QPCh. 6 - Prob. 6.128QPCh. 6 - Prob. 6.129QPCh. 6 - Prob. 6.130QPCh. 6 - Prob. 6.131QPCh. 6 - Prob. 6.132QPCh. 6 - Prob. 6.133QPCh. 6 - Prob. 6.134QPCh. 6 - Prob. 6.135QPCh. 6 - Prob. 6.136QPCh. 6 - Prob. 6.137QPCh. 6 - Prob. 6.138QPCh. 6 - Prob. 6.139QPCh. 6 - Prob. 6.140QPCh. 6 - Prob. 6.141QPCh. 6 - Prob. 6.142QPCh. 6 - Prob. 6.143QPCh. 6 - Prob. 6.144QPCh. 6 - Prob. 6.145APCh. 6 - Prob. 6.146APCh. 6 - Prob. 6.147APCh. 6 - Prob. 6.148APCh. 6 - Prob. 6.149APCh. 6 - Prob. 6.150APCh. 6 - Prob. 6.151APCh. 6 - Prob. 6.152APCh. 6 - Prob. 6.153APCh. 6 - Prob. 6.154APCh. 6 - Prob. 6.155APCh. 6 - Prob. 6.156APCh. 6 - Prob. 6.157APCh. 6 - Prob. 6.158APCh. 6 - Prob. 6.159APCh. 6 - Prob. 6.160APCh. 6 - Prob. 6.161APCh. 6 - Prob. 6.162APCh. 6 - Prob. 6.163APCh. 6 - Prob. 6.164APCh. 6 - Prob. 6.165APCh. 6 - Prob. 6.166APCh. 6 - Prob. 6.167APCh. 6 - Prob. 6.168APCh. 6 - Prob. 6.169APCh. 6 - Prob. 6.170APCh. 6 - Prob. 6.171APCh. 6 - Prob. 6.172APCh. 6 - Prob. 6.173APCh. 6 - Prob. 6.174APCh. 6 - Prob. 6.175APCh. 6 - Prob. 6.176APCh. 6 - Prob. 6.177APCh. 6 - Prob. 6.178APCh. 6 - Prob. 6.179APCh. 6 - Prob. 6.180APCh. 6 - Prob. 6.181APCh. 6 - Prob. 6.182APCh. 6 - Prob. 6.183APCh. 6 - Prob. 6.184APCh. 6 - Prob. 6.185APCh. 6 - Prob. 6.186APCh. 6 - Prob. 6.187APCh. 6 - Prob. 6.188AP
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