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?

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%

Final answer:

The number of moles of MnO4– added to the solution is 0.00106 mol. The moles of Fe2+ in the sample is 0.00530 mol. The grams of iron in the sample is 0.296 g and the percentage of iron in the sample is 33.33%.

Explanation:

The titration reaction between Fe2+ and MnO4– in acid solution is as follows: 5Fe2+ + MnO4– + 8H+ → 5Fe3+ + Mn2+ + 4H2O. Firstly, we can calculate the number of moles of MnO4– added to the solution using the formula volume X molarity. That is 0.0472 L X 0.02240 M = 0.00106 mol MnO4–. According to the balanced redox reaction, one mole of MnO4– reacts with five moles of Fe2+. Therefore, the moles of Fe2+ in the sample is 0.00106 mol MnO4– X 5 = 0.00530 mol Fe2+.

Next, we calculate the grams of iron in the sample. The molarmass of iron is approximately 55.85 g/mol, thus the grams of iron in the sample are 0.00530 mol Fe2+ X 55.85 g/mol = 0.296 g Fe2+. Finally, we find the percentage of iron in the sample by mass = (mass of iron in the sample / total mass of the sample) X 100%. Therefore, the percentage is (0.296 g / 0.8890 g) X 100% = 33.33%.

Learn more about Chemistry and Redox Reactions here:

brainly.com/question/35891199

#SPJ3