percentage

2009-04-08 4:38 am

In the determination of the iron content of an iron ore sample, 4.800g of the sample was weighed out accurately; dissolved in dilute sulphuric acid and the sample solution made up to 500.0cm^3 in a volumetric flask.
A standard solution was prepare by dissolveing 1.1679g of potassium dichromate in 250.0cm^3 of distilled water.
25.0cm^3 of the sample solution was transferred into a conical flask,the iron in the sample solution was first reduced ti iron(II) before titrating with the stardand potassium dichromate solution. it was found that 25.0cm^3 of the sample solution required 20.85cm^3 of the stardand potassium dichromate solution for titration to the end-point.

Calculate the percentage of Fe2O3 in the iron ore sample
(Assuming all ion is in the form of Fe2O3 in one)

thx anyone
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回答 (1)

老爺子

2009-04-08 7:40 am

✔ 最佳答案

Consider the preparation of K2Cr2O7 solution:

Molar mass of K2Cr2O7 = 39x2 + 52x2 + 16x7 = 294 g/mol
Molarity of K2Cr2O7 solution = (1.1679/294)/(250/1000) = 0.01589 M

Consider the titration:
6Fe2+(aq) + Cr2O72-(aq) + 14H+(aq) → 6Fe3+(aq) + 2Cr3+(aq) + 7H2O(l)
Mole ratio Fe2+ : Cr2O7 = 6 : 1

No. of moles of Cr2O72- = 0.01589 x (20.85/1000) = 0.0003313 mol
No. of moles of Fe2+ = 0.0003313 x 6 mol
Molarity of Fe2+ = (0.0003313 x 6)/(25/1000) = 0.07951 M

Consider the preparation of Fe2+(aq) solution.
Fe2O3(aq) → 2Fe3+(aq) → 2Fe2+(aq)
Mole ratio Fe2O3 : Fe2+ = 1 : 2

Total no. of moles of Fe2+ = 0.07951 x (500/1000) = 0.03976 mol
No. of moles of Fe2O3 = 0.03976 x (1/2) = 0.01988 mol
Molar mass of Fe2O3 = 55.9x2 + 16x3 = 159.8 g/mol
Mass of Fe2O3 = 0.01988 x 159.8 = 3.177 g
Mass % of Fe2O3 in the sample = (3.177/4.800) x 100% = 66.19%
=

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