4B Post- + Pre-Lab Answer Keys

# 4B Post- + Pre-Lab Answer Keys - ANSWER KEY EXPERIMENT#4b...

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1 ANSWER KEY: EXPERIMENT #4b - Oxalate Analysis PRE-LAB QUESTIONS: 1. (2 points) The balanced equation: 2 MnO 4 + 16 H + + 10 e 2 Mn 2+ + 8 H 2 O 5 H 2 C 2 O 4 10 CO 2 + 10 e + 10 H + 2 MnO 4 + 5 H 2 C 2 O 4 + 6 H + 2 Mn 2+ + 10 CO 2 + 8 H 2 O Thus, the molar ratio of KMnO 4 : C 2 O 4 2– is 2:5. 2. (2 points) The number of moles of Na 2 C 2 O 4 is: 0.8273 g/134.00 g/mol = 0.0061738 mol of Na 2 C 2 O 4 Thus, the number of moles of KMnO 4 , from the equation in #1, is: 0.0061738 mol × 2/5 = 0.0024695 mol of KMnO 4 Molarity is mol/L, so: 0.0024965 mol/0.02514 L = 0.09823 M KMnO 4 3. (2 points) Again, from the equation in #1, the # of moles of KMnO 4 is: 0.01179 L × 0.02396 Mol/L = 0.0002824 mol of KMnO 4 Again, the molar ratio of C 2 O 4 2- : KMnO 4 is 5:2. 0.0002824 mol × 5/2 = 0.0007062 mol = 0.7062 mmol of C 2 O 4 2- 4. (2 points) So, for mmol of oxalate in 100.0 g of Fe(C 2 O 4 ) 2 , just set up a ratio: sample g 100.0 x sample g 0.1084 mol 0.0007062 = x = 0.6515 mol = 651.5 mmol of C 2 O 4 2- in 100g Carrying on from #3, above, the molecular weight of C 2 O 4 2- (in Fe(C 2 O 4 ) 2 ) is 88.02 g/mol. So: 0.6515 mol × 88.02 g/mol = 57.34 g of C

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4B Post- + Pre-Lab Answer Keys - ANSWER KEY EXPERIMENT#4b...

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