Question

If 100.0 mL of 0.443 M Na₂SO₄ are added to 100.0 mL of 0.983 M Pb(NO₃)₂, how many grams of PbSO₄ can be produced?

Na₂SO₄(aq) + Pb(NO₃)₂(aq) → 2NaNO₃(aq) + PbSO₄(s)

Answer 1

To determine how many grams of PbSO4 can be produced, we calculate the moles of each reactant, identify the limiting reactant, and use its moles with the molar mass of PbSO4 to find the mass of PbSO4 produced, which is 13.44 grams.

Step-by-step explanation:

When 100.0 mL of 0.443 M Na2SO4 are added to 100.0 mL of 0.983 M Pb(NO3)2, we can calculate the amount of PbSO4 that can be produced using stoichiometry. To find the limiting reactant, we first calculate the moles of each reactant. For Na2SO4: 0.100 L × 0.443 mol/L = 0.0443 moles. For Pb(NO3)2: 0.100 L × 0.983 mol/L = 0.0983 moles. However, because the reaction requires one mole of Na2SO4 for each mole of Pb(NO3)2, Na2SO4 is the limiting reactant. Using the molar mass of PbSO4 (303.26 g/mol), we can now calculate the mass of precipitate formed: 0.0443 moles of Na2SO4 will produce an equal amount of PbSO4 since it's a 1:1 ratio, resulting in 0.0443 moles × 303.26 g/mol = 13.44 grams of PbSO4.