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A 1.45 g sample of an iron ore is dissolved in an acid and the iron is obtained as Fe+2 (ag). To titrate the solution, 21.6 mL of 0.102 M KMnO4 (aq) is required. The Fe+2 solution reacts with MnO4- as follows: 5Fe+2 (aq) + MnO-4 (aq) + 8H+ (aq) --------> 5Fe+3 (aq) + Mn+2 (aq) 4H2O a. What is the percent of iron in the ore?

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IthaloAbreu

Answer:

42.3%

Explanation:

By the reaction given the stoichiometry between MnO₄⁻ and Fe⁺² is 1 mol of MnO₄⁻ to 5 moles of Fe⁺². When KMnO₄ dissolves, it forms the same amount of K⁺ and MnO₄⁻ (1:1:1), so the number of moles of MnO₄⁻ used is:

n = Volume*concentration

n = 0.0216 L * 0.102 mol/L

n = 2.20x10⁻³ mol

1 mol of MnO₄⁻ -------------------- 5 moles of Fe⁺

2.20x10⁻³ mol ---------------------- x

By a simple direct three rule

x = 0.011 mole

The molar mass of iron is 55.8 g/mol. The mass is the molar mass multiplied by the number of moles, thus:

m = 55.8*0.011

m = 0.614 g

Then, the percent of iron in the ore is:

(mass of iron/ mass of ore) *100%

(0.614/1.45)*100%

42.3%

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