World of Chemistry
World of Chemistry
7th Edition
ISBN: 9780618562763
Author: Steven S. Zumdahl
Publisher: Houghton Mifflin College Div
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Chapter 9, Problem 20A
Interpretation Introduction

Interpretation:

The amount of iron (III) chloride formed when 15.5 mg of iron reacts with excess chlorine gas 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 number 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 20A

Mass of FeCl3 = 44.9 mg i.e. 45.0 mg

Explanation of Solution

The given reaction is:

  Fe(s) + Cl2(g) FeCl3(s) 

The balanced equation is:

  2Fe(s) + 3Cl2(g) 2FeCl3(s) 

Here since Cl2 is present in excess and Fe is the limiting reagent. The amount of FeCl3 formed will be determined by the amount of Fe reacted.

Step 1: Calculate the moles of Fe present:

Mass of Fe present = 15.5mg = 0.0155 g

Atomic weight of Fe = 55.85 g/mole

  Moles of Fe = Mass of FeMolar Mass Fe=0.0155 g55.85 g/mol=0.000277 moles

Step 2: Calculate the moles of FeCl3 formed:

Based on the reaction stoichiometry:

Fe and FeCl3 are present in a 1:1 molar ratio

Therefore, 0.000277 moles of Fe would produce 0.000277 moles of FeCl3

Step 3: Calculate the mass of FeCl3 formed:

Moles of FeCl3 formed= 0.000277 moles

Molecular weight of FeCl3 = 162.2 g/mol

  Mass of FeCl3 = Moles × Molecular weight=0.000277 moles × 162.2 g/mol = 0.0449 g

Conclusion

Therefore, mass of FeCl3 formed is around 44.9 mg.

Chapter 9 Solutions

World of Chemistry

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