Chem Stoichiometry2(10 PTS)

2009-09-26 5:05 pm

1.) Manganate(VII) solution oxidize iron(II) ions to
iron(III) according to the following ionic equation:

5Fe^2+ + MnO4^- + 8H^+ ----> 5Fe^3+ +Mn^2+ 4H2O

2.5g of potassium manganate (VII) was dissolved in 50.0cm^3
of 0.1 M sulphuric acid . Distilled water was then added to the
mixture until the solution was made up to 250.0 cm^3 in a
volumetric flask. At the same time,4.5g of an iron(II) salt was
dissolved in distilled water and the solution was made up to
100.0cm^3.
25.0cm^3 of the acidified manganate (VII) solution required
33.4cm^3 of this iron(II) salt solution for complete reaction.

a.) Calculate the molarity of the acidified potassium
manganate (VII) solution.

b.)From the results,calculate the molarity of the iron(II) solution.

c.) Given the molar mass of the iron(II) salt is 152.3g mol^-1,
what is hte percentage purity of the salt?

Thank you so much!!

回答 (1)

六呎將軍

2009-09-26 5:47 pm

✔ 最佳答案

(a) No. of moles of KMnO4 = 2.5/158 = 0.0158
So molarity of KMnO4 = 0.0158/0.25 = 0.0633 M
(b) According to the equation, 1 mole of MnO4- reacts completely with 5 moles of Fe2+ and hence:
No. of moles of MnO4- = 0.0633 x 0.025 = 0.00158
No. of moles of Fe2+ = 0.00158 x 5 = 0.00791
Thus molarity of Fe2+ = 0.00791/0.0334 = 0.237 M
(c) No. of moles of Fe2+ present in 100 cm3 of the iron(II) salt solution = 0.237 x 0.1 = 0.0237
Hence mass of the iron(II) salt present in the original 4.5 g of sample = 0.0237 x 152.3 = 3.61 g
Finally % purity = 3.61/4.5 x 100% = 80.17%

參考: Myself

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