World of Chemistry, 3rd edition
World of Chemistry, 3rd edition
3rd Edition
ISBN: 9781133109655
Author: Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher: Brooks / Cole / Cengage Learning
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Chapter 9, Problem 24A
Interpretation Introduction

Interpretation:

The amount of nitrogen, oxygen and water vapor formed when 1.25 g of ammonium nitrate decomposes needs to be deduced based on the given reaction.

Concept Introduction:

  • A chemical reaction is expressed as a chemical equation having reactants and products on left and right side of the reaction arrow respectively.
  •   Reactants  Products

  • The coefficient of a balanced chemical equation i.e. the stoichiometry gives the amount of reactants and products involved in the reaction.
  • Chemical equations can therefore be used to determine the amount of products formed from a known quantity of reactants.

Expert Solution & Answer
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Answer to Problem 24A

Mass of N2 = 0.437 g, Mass of O2 = 0.499 g , Mass of water vapour =0.562 g.

Explanation of Solution

The given reaction is:

  NH4NO3(s) N2(g) + O2(g) + H2O(g)

The balanced equation is:

  NH4NO3(s) N2(g) + O2(g) + 2H2O(g)

Step 1: Calculate the moles of NH4NO3 present:

Mass of NH4NO3 present= 1.25 g

Molar mass of NH4NO3 = 80 g/mole

  Moles of NH4NO3Mass of NH4NO3Molar Mass NH4NO3=1.25 g80 g/mol=0.0156 moles

Step 2: Calculate the mass of N2 formed:

Based on the reaction stoichiometry:

1 mole of ammonium nitrate forms 1 mole of N2 gas

Therefore, 0.0156 moles of ammonium nitrate would yield 0.0156 moles of N2.

Molecular weight of N2 = 28 g/mol

  Mass of N2 = Moles × Molecular weight=0.0156 moles × 28 g/mol = 0.437 g

Step 3: Calculate the mass of O2 formed:

Based on the reaction stoichiometry:

1 mole of ammonium nitrate forms 1 mole of O2 gas

Therefore, 0.0156 moles of ammonium nitrate would yield 0.0156 moles of O2

Molecular weight of O2 = 32 g/mol

  Mass of O2 = Moles × Molecular weight=0.0156 moles × 32 g/mol = 0.499 g

Step 4: Calculate the mass of water vapor formed:

Based on the reaction stoichiometry:

1 mole of ammonium nitrate forms 2 moles of H2O gas

Therefore, 0.0156 moles of ammonium nitrate would yield:

  0.0156 moles NH4NO3×2 moles H2O1 mole NH4NO3=0.0312 moles H2O

Molecular weight of H2O = 18 g/mol

  Mass of H2O = Moles × Molecular weight=0.0312 moles × 18 g/mol = 0.562 g

Conclusion

Therefore, the amount of nitrogen, oxygen and water vapor formed are 0.437 g, 0.499 g and 0.562 g respectively.

Chapter 9 Solutions

World of Chemistry, 3rd edition

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