Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN: 9781285199023
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Textbook Question
Chapter 4, Problem 4.26PAE

4.24 Ammonia gas can be prepared by the reaction

   CaO ( s )   +  2 NH 4 Cl ( s )  2 NH 3 ( g )   +  H 2 O ( g )   +  CaCl 2 ( s )

If 112 g of CaO reacts with 224 g of NH4Cl, how many moles of reactants and products are there when the reaction is complete?

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation:

To calculate how many moles of reactants and products will be left after the completion of reaction.

Concept introduction:

  • Reactants decrease in moles, Products increase in moles.
  • Stoichiometry relates all species to their atomic ratios.
  • Limiting reactant will never be leftover if 100 % completion.
  • Excess reactant will always be leftover.

Given:

= 112 g of CaO= 224 g of NH4Cl

Answer to Problem 4.26PAE

Solution:

Reactatns:0.20molNH4ClProducts:4.0molNH32.0molH2O2.0molCaCl2

Explanation of Solution

When a reaction occurs, there are reactants reacting in order to form new products. The reactants are not typically added in a 100 % stoichiometric ratio, therefore, there is always a limiting and an excess reactant.

It is much easier to convert to molar units, rather than mass units, in order to verify the reactants and products present.

Step 1: Balance Reaction

The reaction is already balanced

Step 2: Change all mass units to mol units

MolecularWeight(MW) CaO = 56.0774g/mol

MolecularWeight(MW) of NH4Cl = 53.491g/mol

Moles of CaO= mass of CaO MW of CaO=  112 g 56.0774 g/mol  = 2.0molMoles of NH4Cl= mass of N H 4 Cl MW of N H 4 Cl= 224 g 53.491g/mol=4.20mol

Step 3: Relate stoichiometry

1 mol of CaO : 2 mol of NH4Cl : 2 mol of NH3 : 1 mol of H2O : 1 mol of CaCl2

Step 4: Moles reacted

Identify the limiting reactant ( CaO

molCaOleft=2.02.0=0.0

Excess reactant of NH4Cl

molNH4Cl=4.20(2molNH4Cl1molCaO)(2.0molCaO)=0.20molNH4Cl

Step 5: Moles produced

Moles of product NH3 formed

molNH3=0+(2molNH31molCaO)(2.0molCaO)=4.0molNH3

Moles of product H2O H2O formed

molH2O=0+(1molH2O1molCaO)(2.0molCaO)=2.0molH2O

Moles of product CaCl2 formed

molCaCl=20+(1molCaCl21molCaO)(2.0molCaO)=2.0molCaCl2

Conclusion

Use stoichiometry to verify the ratio between species, that is, reactants and products after the reaction is over.

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Chapter 4 Solutions

Chemistry for Engineering Students

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