Question

A 1.555 g mixture of the solid salts Na2SO4 (MM=142.04) + Pb(NO3)2 (MM=331.21) forms an aqueous solution with the precipitation of PbSO4 (MM=303.26). The precipitate was filtered and dried, and its mass was determined to be 0.68 g. The limiting reactant was determined to be Na2SO4. Determine the mass in g of Pb(NO3)2 in the original salt mixture. Na2SO4 (aq) + Pb(NO3)2 (aq) -> PbSO4 (s) + 2 NaNO3 (aq)

Answer 1

Na2SO4 (aq) + Pb(NO3)2 (aq) -> PbSO4 (s) + 2 NaNO3 (aq)

m = 1.555 g of salts

mass of precipitate, PbSO4 = 0.68 g

mol of PbSO4 = mass/MW = 0.68/303.26 = 0.002242 mol of PbSO4

limiting was Na2SO4...

balance:

mass of Na2SO4 + mass of Pb(NO3)2 = 1.555

mol of Na2SO4 + mol of Pb(NO3)2 = 0.002242

we know that ratio is:

1 mol of Na2SO4 = 1 mol of PbSO4

0.002242 mol of PBSO4 = 0.002242 mol of Na2SO4

then

mass of Na2SO4 = 0.002242*142.04 = 0.318453 g of Na2SO4

then

mass of PB(NO3)2 = Total mass - Mass of NA2SO4 = 1.555-0.318453 = 1.236547 g of PB(NO3)2