Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 6, Problem 6.82QP
Interpretation Introduction

Interpretation: The amount of potassium nitrate and potassium chlorate needed to make 200.0L of gas at 1.00atm and 290K is to be calculated.

Concept introduction: The thermal decomposition of potassium chlorate generates oxygen. Nitrogen is produced from the reaction of sodium metal and potassium nitrate.

To determine: The amount of potassium nitrate and potassium chlorate needed to make 200.0L of gas at 1.00atm and 290K .

Expert Solution & Answer
Check Mark

Answer to Problem 6.82QP

Solution

The amount of KNO3(s) required is 1.358×103g_ and the amount of KClO3(s) required is 1.37×102g_ .

Explanation of Solution

Explanation

Given

The pressure is 1.00atm .

The temperature is 290K .

The volume of gas is 200.0L .

The volume of N2 is 160L .

The volume of O2 is 42L .

The amount of N2 ( n ) is calculated as,

n=PVRT

Where,

  • P is the pressure.
  • V is the volume of N2 .
  • R is gas constant (0.08206l.atm/mol.K)
  • T is temperature.

Substitute the value of P , V , R and T in the above equation.

n=1.00atm×160L0.08206Latm/molK×290K=6.72mol

The amount of O2 ( n1 ) is calculated as,

n1=PVRT

Where,

  • P is the pressure.
  • V is the volume of N2 .
  • R is gas constant (0.08206l.atm/mol.K)
  • T is temperature.

Substitute the value of P , V , R and T in the above equation.

n=1.00atm×40L0.08206Latm/molK×290K=1.68mol

The chemical equation for the reaction of sodium metal and potassium nitrate to generate nitrogen is,

10Na(s)+2KNO3(s)K2O(s)+5Na2O(s)+N2(g)

According to the above reaction,

2molKNO3(s)1molN2(g)

The molar mass of KNO3(s) is 101.1g/mol . Therefore, mass of 1molKNO3(s) is equal to 101.1g . Hence mass of 2molKNO3(s) is equal to 2×101.1g=202.2g .

Since 1molN2 is produced from 202.2g of KNO3(s) .

Therefore, 6.72molN2 is produced from 6.72×202.2gKNO3(s)=1.358×103g_KNO3(s)

Therefore, the amount of KNO3(s) required is 1.358×103g_ .

The oxygen is generated by the thermal decomposition of potassium chlorate.

2KClO3(s)2KCl(s)+3O2(g)

According to the above reaction,

2molKClO3(s)3molO2(g)

The molar mass of KClO3(s) is 122.55g/mol . Therefore, mass of

2molKClO3(s)=2×122.55=245.1g

Since 3molO2(g) is produced from 245.1g of KClO3(s) .

Therefore, 1.68molO2 is produced from 245.1×1.683gKClO3(s)=1.37×102g_KClO3(s)

Therefore, the amount of KClO3(s) required is 1.37×102g_ .

Conclusion

The amount of KNO3(s) required is 1.358×103g_ and the amount of KClO3(s) required is 1.37×102g_

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Chapter 6 Solutions

Chemistry

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