A 1.45 g sample of an iron ore is dissolved in acid and the iron obtained is Fe2+(aq). To titrate the solution, 21.6 mL of 0.102 M KMnO4(aq) is required. What is the percent of iron in the ore? Hint: calculate moles of iron then mass of iron then % iron.

42.4%

Explanation:

5Fe2+(aq) + MnO4^- (aq) + 8H^+(aq)> 5Fe^3+(aq) + 4H2O(l) + Mn2+(aq)

Concentration of oxidizing agent =0.102M

Volume of oxidizing agent= 21.6ml

Amount of MnO4^- reacted= CV= 21.6ml/1000 × 0.102= 2.2 × 10^-3moles

From the balanced redox reaction equation

5 moles of Fe^2+ required 1 mole of MnO4^-

x moles of Fe^2+ will require 2.2 ×10^-3 moles of MnO4^-

x= 5×2.2×10^-3/1 = 11 ×10^-3moles of Fe^2+

Therefore, mass of Fe^2+ reacted = 11×10^-3 × 56= 616×10^-3g of Iron

% 0f iron present = 0.616/1.45 × 100= 42.4%

ca has a crystalline lattice with positive ions (since it is a metal, metals are positively charged) and electrons flow freely in its structure since the ca atoms have a large nuclear radius so the electrons are not tightly held, therefore the electrons can flow freely since they are not being highly attracted by the nucleus. moreover, one of the properties of metals are that they are conductors of electricity, thus proving ca is the element.