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%
