Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 6, Problem 6.81QP
Interpretation Introduction

Interpretation: The amount of ammonium dichromate and potassium chlorate needed to make 200.0L of air at 0.85atm and 273K is to be calculated.

Concept introduction: The thermal decomposition of potassium chlorate generates oxygen and decomposition of ammonium dichromate generates pure nitrogen.

To determine: The amount of ammonium dichromate and potassium chlorate needed to make 200.0L of air at 0.85atm and 273K .

Expert Solution & Answer
Check Mark

Answer to Problem 6.81QP

Solution

The amount of (NH4)2Cr2O7(s) required is 1.487×103g_ and the amount of KClO3(s) required is 1.2×102g_ .

Explanation of Solution

Explanation

Given

The pressure is 0.85atm .

The temperature is 273K .

The volume of air is 200.0L .

Air is about 78% nitrogen by volume. Therefore, the volume of nitrogen is calculated as,

200.0L×78100=156L

Therefore, the volume of N2 is 156L .

Air is about 21% oxygen by volume. Therefore, the volume of oxygen is calculated as,

200.0L×21100=42L

Therefore, 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=0.85atm×156L0.08206Latm/molK×273K=5.9mol

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.

n1=0.85atm×42L0.08206Latm/molK×273K=1.5mol

The decomposition reaction of ammonium chromate is,

(NH4)2Cr2O7(s)N2(g)+Cr2O3(s)+4H2O(g)

According to the above reaction,

1mol(NH4)2Cr2O7(s)1molN2(g)

The molar mass of (NH4)2Cr2O7(s) is 252.07g/mol . Therefore, mass of 1mol(NH4)2Cr2O7(s) is equal to 252.07g .

Since 1molN2 is produced from 252.07g of (NH4)2Cr2O7(s) .

Therefore, 5.9molN2 is produced from 252.07×5.9g(NH4)2Cr2O7(s)=1.487×103g_(NH4)2Cr2O7(s)

Therefore, the amount of (NH4)2Cr2O7(s) required is 1.487×103g_ .

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, 1.5molO2 is produced from 245.1×1.53gKClO3(s)=1.2×102g_KClO3(s)

Therefore, the amount of KClO3(s) required is 1.2×102g_ .

Conclusion

The amount of (NH4)2Cr2O7(s) required is 1.487×103g_ and the amount of KClO3(s) required is 1.2×102g_

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Chapter 6 Solutions

Chemistry

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