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A sample of an iron ore is dissolved in acid, and the iron is converted to Fe2+. The sample is then titrated with 47.20 mL of 0.02240 M MnO4– solution. The oxidation-reduction reaction that occurs during titration is(a) How many moles of MnO4– were added to the solution? (b) How many moles of Fe2+ were in the sample?(c) How many grams of iron were in the sample? (d) If the sample had a mass of 0.8890 g, what is the percentage of iron in the sample?

Respuesta :

sidrawasim768

Answer:

33.21% is the right answer.

Explanation:

0.04720 L * 0.02240 mol / L = 1.0573*10-3

mol Fe+2 = 1.0573* 10-3 mol = 5.2864*10-3

mass of Fe= 5.2864 * 10-3  = 0.29522g

% of Fe in sample = 0.29522 g Fe / 0.8890 g of sample * 100 = 33.21%

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