A 0.3147 g sample of primary-standard-grade Na 2 C 2 O 4 was dissolved in H 2 SO

Question

A 0.3147 g sample of primary-standard-grade Na2C2O4was dissolved in H2SO4 and titrated to the end point with 31.67 mL of dilute KMnO4;

2MnO4- + 5C2O42- + 16H+2Mn2+ + 10CO2 (g) + 8H2O

a) Calculate the molar concentration of KMnO4 solution.

b) The iron in a 0.6656 g ore sample as reduced quantitatively to the +2 state and then titrated with 26.75 mL of the KMnO4 solution from part a). The ionic equation is shown below;

5Fe2+ + MnO4- + 8H+5Fe3+ + Mn2+ + 4H2O

Calculate the percent Fe2O3 (FM= 159.69) in the sample.

Explanation / Answer

(a) 2 MnO4- + 5 C2O42- + 16 H+ => 2 Mn2+ + 10 CO2 (g) + 8 H2O(l)

Moles of C2O42- = moles of Na2C2O4 = mass/molar mass of Na2C2O4

= 0.3147/134.00 = 0.0023485 mol


Moles of KMnO4 = moles of MnO4- = 2/5 x moles of C2O42-

= 2/5 x 0.0023485 = 0.00093941 mol


Concentration of KMnO4 = moles/volume of KMnO4

= 0.00093941/0.03167

= 0.029662 M = 0.02966 M


(b) 5 Fe2+ + MnO4- + 8 H+ => 5 Fe3+ + Mn2+ + 4 H2O

Moles of MnO4- = moles of KMnO4 = volume x concentration of KMnO4

= 26.75/1000 x 0.029662 = 0.00079347 mol


Moles of Fe2+ = 5 x moles of MnO4-

= 5 x 0.00079347 = 0.00396735 mol


Moles of Fe2O3 = 1/2 x moles of Fe2+

= 1/2 x 0.00396735 = 0.0019837 mol


Mass of Fe2O3 = moles x molar mass of Fe2O3

= 0.0019837 x 159.69 = 0.31677 g


Mass% of Fe2O3 = mass of Fe2O3/mass of sample x 100%

= 0.31677/0.6656 x 100%

= 47.59%

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