Question

If 100.0 mL of 0.171 M Na2SO4 are added to 100.0 mL of 0.889 M Pb(NO3)2, how many grams of PbSO4 can be produced?

Equation: Na2SO4 + Pb(NO3)2 = 2NaNO3 + PbSO4
mols Na2SO4 = M x L = ?
mols Pb(NO3)2 = M x L = ?
It should be obvious that Na2SO4 is the limiting reagent. 1 mol Na2SO4 produces 1 mol PbSO4; therefore, mols PbSO4 = mols Na2SO4.
Then grams PbSO4 = mols PbSO4 x molar mass PbSO4

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