Chemistry 2012 Student Edition (hard Cover) Grade 11
Chemistry 2012 Student Edition (hard Cover) Grade 11
12th Edition
ISBN: 9780132525763
Author: Prentice Hall
Publisher: Prentice Hall
Question
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Chapter 12, Problem 63A
Interpretation Introduction

Interpretation: To calculate the total number of liters of gas.

Concept Introduction: A mole is 6.02×1023 atoms/molecules. The SI unit for the amount of substance is a mole and it is represented by “mol”.

The coefficients are the numbers that are written in front of the chemical formula/symbol in order to balance a chemical reaction.

Expert Solution & Answer
Check Mark

Answer to Problem 63A

The total number of liters of nitrogen gas is 63.80 L .

Explanation of Solution

The given chemical reaction is depicted as follows:

  2NH4NO3s2 N2g+4 H2Ol+O2g

The decomposition of two moles of ammonium nitrate it yields two moles of nitrogen, four moles of water, and one mole of oxygen.

The amount of ammonium nitrate is 228 g  . At STP, 1 mol of the gas has molar volume 22.4 L.

Calculation for the total number of liters of nitrogen gas is as follows:

  228 g NH4NO3×1mol NH4NO3 80.04 g NH4NO3× 2mol N22mol NH4NO3×22.4 L N2 1mol N2=63.80 L N2

Conclusion

Thus, the total number of liters of nitrogen gas is 63.80 L .

Chapter 12 Solutions

Chemistry 2012 Student Edition (hard Cover) Grade 11

Ch. 12.2 - Prob. 12SPCh. 12.2 - Prob. 13SPCh. 12.2 - Prob. 14SPCh. 12.2 - Prob. 15SPCh. 12.2 - Prob. 16SPCh. 12.2 - Prob. 17SPCh. 12.2 - Prob. 18SPCh. 12.2 - Prob. 19SPCh. 12.2 - Prob. 20SPCh. 12.2 - Prob. 21LCCh. 12.2 - Prob. 22LCCh. 12.2 - Prob. 23LCCh. 12.2 - Prob. 24LCCh. 12.2 - Prob. 25LCCh. 12.3 - Prob. 26SPCh. 12.3 - Prob. 27SPCh. 12.3 - Prob. 28SPCh. 12.3 - Prob. 29SPCh. 12.3 - Prob. 30SPCh. 12.3 - Prob. 31SPCh. 12.3 - Prob. 32SPCh. 12.3 - Prob. 33SPCh. 12.3 - Prob. 34LCCh. 12.3 - Prob. 35LCCh. 12.3 - Prob. 36LCCh. 12.3 - Prob. 37LCCh. 12.3 - Prob. 38LCCh. 12 - Prob. 39ACh. 12 - Prob. 40ACh. 12 - Prob. 41ACh. 12 - Prob. 42ACh. 12 - Prob. 43ACh. 12 - Prob. 44ACh. 12 - Prob. 45ACh. 12 - Prob. 46ACh. 12 - Prob. 47ACh. 12 - Prob. 48ACh. 12 - Prob. 49ACh. 12 - Prob. 50ACh. 12 - Prob. 51ACh. 12 - Prob. 52ACh. 12 - Prob. 53ACh. 12 - Prob. 54ACh. 12 - Prob. 55ACh. 12 - Prob. 56ACh. 12 - Prob. 57ACh. 12 - Prob. 58ACh. 12 - Prob. 59ACh. 12 - Prob. 60ACh. 12 - Prob. 61ACh. 12 - Prob. 62ACh. 12 - Prob. 63ACh. 12 - Prob. 64ACh. 12 - Prob. 65ACh. 12 - Prob. 66ACh. 12 - Prob. 67ACh. 12 - Prob. 68ACh. 12 - Prob. 69ACh. 12 - Prob. 70ACh. 12 - Prob. 71ACh. 12 - Prob. 72ACh. 12 - Prob. 73ACh. 12 - Prob. 74ACh. 12 - Prob. 75ACh. 12 - Prob. 76ACh. 12 - Prob. 77ACh. 12 - Prob. 78ACh. 12 - Prob. 79ACh. 12 - Prob. 80ACh. 12 - Prob. 81ACh. 12 - Prob. 82ACh. 12 - Prob. 83ACh. 12 - Prob. 84ACh. 12 - Prob. 85ACh. 12 - Prob. 88ACh. 12 - Prob. 89ACh. 12 - Prob. 90ACh. 12 - Prob. 91ACh. 12 - Prob. 92ACh. 12 - Prob. 93ACh. 12 - Prob. 94ACh. 12 - Prob. 95ACh. 12 - Prob. 96ACh. 12 - Prob. 97ACh. 12 - Prob. 98ACh. 12 - Prob. 99ACh. 12 - Prob. 100ACh. 12 - Prob. 101ACh. 12 - Prob. 102ACh. 12 - Prob. 103ACh. 12 - Prob. 104ACh. 12 - Prob. 105ACh. 12 - Prob. 106ACh. 12 - Prob. 107ACh. 12 - Prob. 1STPCh. 12 - Prob. 2STPCh. 12 - Prob. 3STPCh. 12 - Prob. 4STPCh. 12 - Prob. 5STPCh. 12 - Prob. 6STPCh. 12 - Prob. 7STPCh. 12 - Prob. 8STPCh. 12 - Prob. 9STPCh. 12 - Prob. 10STP
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