Question

3 attempts left Check my work Enter your answer in the provided box. A chemical engineer determines the mass percent of iron in an ore sample by converting the Fe to Feat in acid and then titrating the Fe?* with MnO4 A 1.0986-g sample was dissolved in acid and then titrated with 31.36 mL of 0.04891 M KMnO4. The balanced equation is 8 H*(aq) + 5 Fe2+ (aq)+ Mno, (aq) → 5 Fe3+(aq) + Mn²+(aq) + 4 H20(1) Calculate the mass percent of iron in the ore. I % Fe

Answer 1

Given

Volume of MnO₄^- = 31.36 mL = 0.03136 L

Concentration of MnO₄^- (Molarity) = 0.04891 M

Molarity = moles/ litre

moles = 0.04891 M x 0.03136 L

moles = 0.0015357 moles of MnO^4-

from balanced equation we know 1 mol of MnO^4- was used to converts 5 moles of Fe²⁺

So moles of Fe²⁺ = 5 x moles of MnO^4-

moles of Fe²⁺ = 5 x 0.0015357 = 0.007675 moles of Fe²⁺

molar mass of Fe = 55.85 g/mol

mass = moles x molar mass

mass of Fe = 0.007675 moles x 55.85 g/mol

mass of Fe = 0.4286 g

% Fe in ore = mass of Fe / mass of ore x 100

% Fe in ore =  0.4286 g / 1.0986 g x 100

% Fe in ore = 39.01 %