Asample 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?

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%

thats how its supposed to be

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3