Letter Expl 2

Dear Lynn L. Brown,

We have evaluated your iron ore deposit samples that have been converted into iron sulfate (FeSO₄). To find out the amount of iron in this slurry, we performed a titration.

We began this process by cleaning a 50 mL buret with water, rinsing it 3 times. We then filled 50 mL of 0.02 M KMnO₄ in a beaker and 15 mL of FeSO₄-7H₂O solution in another. We poured the KMnO₄ into the buret and emptied any extra to get it to the 50 mL mark on the buret. To have a color standard, you can fill a flask with 50 mL of distilled water, add a drop of 0.02 M KMnO₄, and set aside. With the 15 mL of FeSO₄-7H₂O, we placed it under the buret and slowly released the KMnO₄ while stirring. We stopped the titration once it reached a light orange color. We recorded our volume of KMnO4 used in our titration and repreated hte experiment twice more.

To get the concentration of FeSO₄ concentration, we took the average of both the volumes of KMnO₄ and FeSO₄ and the concentration of KMnO₄ and found each average. We used the equation C_AV_A = C_BV_B. Therefore, we multiplied the concentration and volume of KMnOr and divided by the volume of FeSO₄-7H₂O to find its concentration, which was 0.0093 M. To go from mol/L to mol/ 10⁵ L of FeSO₄-7H₂O, we multiplied 0.0093 by 105 to get 930. We took the molar mass of Fe and divided it by the molar mass of FeSO₄-7H₂O and found the percentage composition by mass of iron in FeSO₄-7H₂O. It was 20%. We had to multiply 20% by 930 to get the mol/ 10⁵ L of iron (Fe), which was 186. Then we took the 186 mol/ 10⁵ L and multiplied it by molar mass of Fe to find grams of Fe in 10⁵ L, which was 1.04 x 10⁴ g. Then we divided by 1000 and our answer was 1.04 x 10¹ or 10.4 kg/10⁵ L.

We do not recommend you mine from the site because if you need at least 150 kg of FeSO₄ from 1.0 x 10⁵ L of the slurry, there will not be enough.

Sincerely,

Sid the Sloth