When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed lead sulfate precipitates out of soultion. Calculate the mass of lead (II) sulfate that should form when 1.25 L of 0.0500 M Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are reacted together.

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  1. Pb(NO3)2+ Na2SO4 ==> PbSO4 + 2NaNO3
    moles Pb(NO3)2 = 1.25L x 0.0500M = 0.0625.
    moles Na2SO4 = 2L x 0.0250 = 0.0500.
    Since 1 mol Pb(NO3)2 uses 1 mol Na2SO4 to form 1 mol PbSO4, mol PbSO4 formed must be 0.0500. There isn't enough Na2SO4 for all of the Pb(NO3)2 to be used.
    Convert 0.0500 mols PbSO4 to grams. g= mols x molar mass = ?

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  2. 15.2

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  3. 15.1633 g PbSO4

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  4. 15.2g

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