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.

Answer 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 = ?

Answer 2

15.2

Answer 3

15.1633 g PbSO4

Answer 4

15.2g