Titration of a 0.1527 gram sample of the iron-oxalate complex required, after co

Question

Titration of a 0.1527 gram sample of the iron-oxalate complex required, after correction for a blank titration, 20.24 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 1.914E-2 M.

How many moles of potassium permanganate were required?

Moles of KMnO4 = ___________ mol

How many moles of oxalate are present in the sample, and therefore what mass of oxalate is present in the sample?

Moles of oxalate = ______________ mol

Mass of oxalate = ______________ grams

What is the percentage of oxalate in the iron-oxalate complex?

Percent by mass of oxalate in the iron-oxalate complex =__________________ %

Explanation / Answer

5C2O4^2-(aq) + 2MnO4^-(aq) + 16H^+(aq) --> 2Mn^2+(aq) + 10CO2(g) + 8H2O(l)

moles of permanganete = MV = 0.01914 x 20.24/1000 = 0.0003874

moles of oxalte = ( 5/2) x 0.0003874 = 0.00097

mass of oxalte = 0.00097 x 143.91 = 0.1394 gm

mass % = 100 x ( 0.1394/0.1527) = 91.27 %

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.