Answer:
Mass of precipitate (PbSO₄) formed = 0.696 g
Explanation:
The reaction of Lead II Nitrate and Sodium Sulfate is given as
Pb(NO₃)₂ (aq) + Na₂SO₄ (aq) → 2NaNO₃ (aq) + PbSO₄ (s)
The precipitate from this reaction is the PbSO₄
0.76 g of Pb(NO₃)₂ is available for reaction, we convert to number of moles
Number of moles = (Mass)/(Molar Mass)
Mass = 0.76 g
Molar mass of Pb(NO₃)₂ = 331.2 g/mol
Number of moles of Pb(NO₃)₂ = (0.76/331.2) = 0.002294686 moles = 0.002295 moles
25.00 ml of 0.2010M sodium sulfate is available for the reaction, we also conert to number of moles.
Number of moles = (Concentration in mol/L) × (Volume in L)
Concentration of Na₂SO₄ in mol/L = 0.201 M
Volume of Na₂SO₄ in L = (25/1000) = 0.025 L
Number of moles of Na₂SO₄ = 0.201 × 0.025 = 0.005025 moles
From the stoichiometric balance of the reaction, 1 mole of Pb(NO₃)₂ reacts with 1 mole of Na₂SO₄
Hence, this indicates that Pb(NO₃)₂ is the limiting reagent as it is in short supply and it will therefore dictate the amount of precipitate (PbSO₄) formed.
1 mole of Pb(NO₃)₂ gives 1 mole of the precipitate (PbSO₄)
0.002295 mole of Pb(NO₃)₂ will give 0.002295 mole of the precipitate (PbSO₄)
Mass = (Number of moles) × (Molar mass)
Number of moles of precipitate (PbSO₄) formed = 0.002295 mole
Molar mass of PbSO₄ = 303.26 g/mol
Mass of precipitate (PbSO₄) formed = 0.002295 × 303.26 = 0.6958864734 = 0.696 g
Hope this Helps!!!
