Question

Given the following data, determine the % by mass of oxalate (C2O42-) in a sample of an iron oxalate complex with the general formula Kz[Fex(C2O4)y].wH2O.

The reaction between oxalate and permanganate is as follows

: 6H+ + 5(COOH)2 + 2MnO4- → 10CO2 + 2Mn2+ + 8H2O

m(iron oxalate complex) used for titration = 0.100 g

C(KMnO4) standard solution = 0.0200 M

M(C2O42-) = 88.02 g mol-1

Titre volume = 11.91 mL

Enter your answer to 3 significant figures. Do not enter units. Very large or small numbers can be entered in scientific notation - for example enter 1.23E-4 to represent 1.23x10-4. Answer:__________

Question 7- Given the following data, determine the % by mass of Iron (III) (Fe3+) in a sample of an iron oxalate complex with the general formula Kz[Fex(C2O4)y].wH2O.

m(iron oxalate complex) used for stock sample = 0.0100 g

Vtotal (stock sample solution) = 25.00 mL

Valiquot (stock sample solution) = 1.00 mL

Vtotal (diluted sample solution) = 25.00 mL

Cdil(Fe3+) = 1.793 mg L-1 (from calibration curve) Enter your answer to 3 significant figures. Do not include units.

Answer: ___________

Question 8- Given the above data, determine the mole ratio of iron (III) to oxalate (Fe3+:C2O42-) in the sample of iron oxalate complex with the general formula Kz[Fex(C2O4)y].wH2O.

M(Fe3+) = 55.85 g mol-1

M(C2O42-) = 88.02 g mol-1

Round your answer to the nearest whole number. Do not multiply your answers.

Ratio (C2O42-) = n(C2O42-)/ n(Fe3+) = _____ = y

Ratio (Fe3+) = n(Fe3+)/ n(Fe3+) = ______ = x

Therefore molar ratio of Fe3+:C2O42- = _____: _____ = x : y

[Fex(C2O4)y] is an anionic species, whereas the overall complex Kz[Fex(C2O4)y] is neutral. The negative charge of the complex is balanced by the positive charge from the K+. Therefore, what is the stoichiometric ratio of potassium (K+) in the sample? Ratio (K+) = ________ = z

What is the formula for the complex Kz[Fex(C2O4)y]?

Answer 1

Sol :-

From the formula of Molarity :

Molarity of the Solution = Number of moles of solute / Volume solution in L

So,

Number of moles of KMnO₄ or MnO₄^- = Molarity x Volume in L

= 0.0200 M x 0.01191 L

= 0.0002382 mol

Given balanced chemical equation is :

6 H⁺ + 5( COOH)₂ + 2 MnO₄^- -----> 10 CO₂ + 2 Mn²⁺ + 8 H₂O

From equation It is cleared that,

2 moles of MnO₄^- oxidize = 5 moles of C₂O₄^2- or (COOH)₂

So,

0.0002382 moles of MnO₄^- oxidize = 5 x 0.0002382 /2 = 0.0005955 mol of C₂O₄^2-

So,

Mass of C₂O₄^2- = Moles of C₂O₄^2- x Gram molar mass of C₂O₄^2-

= 0.0005955 mol x 88.02 g/mol

= 0.05241591 g

So,

Percentage of C₂O₄^2- = Mass of C₂O₄^2- x 100 / Mass taken

= 0.05241591 g x 100 / 0.100 g

= 52.4 %

Hence, Percentage of C₂O₄^2- in sample = 52.4 %