Question

5 pos. 1. An excess of sodium sulfate was added to a 500. mL sample of polluted water. The

mass of lead (II) sulfate that precipitatcd was 308.88 mg. Determine the mass of lead that was in
the polluted water.
Na2SO4(aq) + Pb2+ (aq) → 2Na(aq) + PbSO4(s)

Answer 1

`m_(Pb^(2+))=0.211gPb^(2+)`

Step-by-step explanation:

Hello!

In this case, according to the stoichiometry of the reaction, it is possible to evidence the 1:1 mole ratio between lead (II) ions and lead (II) sulfate precipitate; that is why we can compute the mass of lead (II) in the polluted water as shown below:

`m_(Pb^(2+))=308.88mgPbSO4*(1gPbSO4)/(1000mgPbSO4) *(207.2gPb^(2+))/(303.26gPbSO4) \\\\m_(Pb^(2+))=0.211gPb^(2+)`

Best regards!