A 0.3708 gram sample of ore contains Pb 3 O 4 and inert material. The lead is pr

Question

A 0.3708 gram sample of ore contains Pb3O4 and inert material. The lead is precipitated as PbSO4 and weighs 0.3699 grams. What is the % w/w Pb composition of the ore sample. Express your answer as % of Pb3O4 in the original sample.
Assumptions: the Pb3O4 reacts completely to form precipitate when H2SO4 is applied to it.
You will need to write a chemical equation that is balanced in terms of the Pb atoms to determine the stoichiometry of the precipitate. (you can safely ignore the balancing of the sulfur and oxygen)

Explanation / Answer

Pb3O4 is made up of 2 molecules of PbO and 1 of PbO2.

It can be represented as: PbO2.2PbO

These two parts react differently according to the following reactions:

2PbO2 + 2H2SO4 2PbSO4 + 2H2O + O2

PbO + H2SO4 PbSO4 + H2O

Multiplying second equation by 4 and adding it to 1, we get:

2(PbO2.2PbO) + 6H2SO4 6PbSO4 + 6H2O + O2

OR

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

So, for 6 moles of ppt, 2 moles of Pb3O4 are needed.

Moles of ppt formed = Mass/MW = 0.3699/303.26 = 0.001219

So,

Moles of Pb3O4 present in the ore = 0.001219/3 = 0.0004063

So, Mass of Pb3O4 in the ore = 0.0004063*MW = 0.0004063*685.6 = 0.2785 g

So,

% of Pb3O4 in the original sample = (0.2785/0.3708)*100 = 75.107 %

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