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[CHEM 1120] LAB4:Preparation and Purification of an Iron - Containing Salt $6.49   Add to cart
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[CHEM 1120] LAB4:Preparation and Purification of an Iron - Containing Salt
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  • Course
  • CHEM 1220 (1220)
  • Institution
  • Langara College ( )
  • Book
  • Principles of Chemistry: A Molecular Approach, Global Edition

[CHEM 1120] LAB #4: This will provide the basis for the prelab and questions and calculations. It is a very short lab with just the 1 page. Received 100%.

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Document information

  • November 17, 2021
  • 1
  • 2021/2022
  • Class notes
  • Kevin bozek
  • All classes

Subjects

  • iron containing salt
  • theoretical yield
  • percentage yield
  • percent yield
  • potassium oxalate
  • how to calculate per
  • ferric nitrate monohydrate
  • how to find theoretical yield
  • how to calculate theoretical yield

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Your Name Date lab done




LAB #4 : Preparation and Purification of an Iron-Containing Salt


DATA
Mass of potassium oxalate _____2.253______g
Mass of ferric nitrate monohydrate _____1.500______g




Mass of empty glass beaker _____51.125_____g
Mass of beaker + K3Fe(C2O4)3 _____52.444_____g
Mass of K3Fe(C2O4)3 _____1.319______g




CALCULATIONS
1. Theoretical Yield:

Fe(NO3)3 • 9H2O + 3K2C2O4 •H2O → K3Fe (C2O4)3•3H2O + 3KNO3 + 9H2O

1 mol Fe(NO3)3 • 9H2O 1 mol K3Fe (C2O4)3•3H2O
a) 1.500 g Fe(NO3)3 • 9H2O x 404.0 g
x 1 mol Fe(NO3)3 • 9H2O
= 0.003713 mol

491.2 g K3Fe (C2O4)3•3H2O
b) 0.03713 mol K3Fe (C2O4)3 x = 1.824 g
1 mol

Since Fe(NO3)3 • 9H2O is the limiting reagent, the theoretical yield is 1.824 g of K3Fe (C2O4)3•3H2O.


2. Percent Yield :


actual yield from experiment 1.319 g
% = x 100% = x 100% = 72.31%
theoretical yield 1.824 g

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