***MUST SHOW WORK TO ALL PARTS WITH MY NUMBERS FOR POINTS*** 1. Titration of a 0
ID: 842538 • Letter: #
Question
***MUST SHOW WORK TO ALL PARTS WITH MY NUMBERS FOR POINTS***
1. Titration of a 0.1417 gram sample of the iron-oxalate complex required, after correction for a blank titration, 16.03 mL of the standardized potassium permanganate solution to reach the end point. The molarity of the potassium permanganate solution had aready been established to be 2.129E-2 M.
How many moles of potassium permanganate were required?
2. How many moles of oxalate are present in the sample?
3. What mass of oxalate is present in the sample?
4. What is the percentage of oxalate in the iron-oxalate complex? (Percent by mass)
***MUST SHOW WORK TO ALL PARTS WITH MY NUMBERS FOR POINTS***
Explanation / Answer
Oxalate is titrated acc to reaction.
2MnO4-+5C2O42-+16H+ --- > 10CO2 +2Mn2+ +8H2O
1. Moles of Potassium permanganate required=Molarity of solution*volume in ltrs
=2.129*10-2*0.01603
=3.413*10-4 mol
2. Acc to balanced equation above.
Moles of oxalate=(5/2)*moles of permanganate=(5/2)*3.413*10-4
=8.532*10-4 mol
3. Mass of oxalate= Moles*mol wt
=8.532*10-4*88
=7.508*10-2 g
3. Mass percent of oxalate= (mass of oxalate/mass of complex)*100
=(7.508*10-2/0.1417)*100
=52.99%
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