Question

A 0.4498 gram sample of ore contains Pb3O4 and inert material. The lead is precipitated as PbSO4 and weighs 0.3413 grams. What is the % w/w Pb3O4 composition of the ore sample. Assumptions: the Pb3O4 reacts completely to form precipitate when H2SO4 is applied to it. Express your answer as % of Pb3O4 in the original sample.

Answer 1

The balanced reaction will be

2Pb3O4 + 6H2SO4 ------------------- 6PbSO4 + 6H2O + O2

Molar mass of PbSO4 = Molar mass of Pb + Molar mass of S + 4 * Molar mass of S

=> 207.2 + 32.06 + 4 * 15.9994

=> 303.25 gm/mol

Number of moles of PbSO4 = Mass/molar mass = 0.3413/303.25 = 1.125 * 10^(-3) moles

Hence, number of moles of Pb3O4 = 1/3 * number of moles of PbSO4 = 1/3 * 1.125 * 10^(-3) = 3.751 * 10^(-4)

Molar mass of Pb3O4 = 3 * molar mass of Pb + 4 * molar mass of O = 685.6 gm

Mass of Pb3O4 = number of moles * molar mass = 3.751 * 10^(-4) mol * 685.6 gm/mol = 0.2571 gms

Mass % of Pb3O4 in sample = Mass of Pb3O4/Mass of sample * 100 = 0.2571/0.4498 * 100 = 57.16%