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

Answer:

$m_{Pb^{2+}}=0.211gPb^{2+}$

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:

$m_{Pb^{2+}}=308.88mgPbSO4*\frac{1gPbSO4}{1000mgPbSO4} *\frac{207.2gPb^{2+}}{303.26gPbSO4} \\\\m_{Pb^{2+}}=0.211gPb^{2+}$

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