0.76 g of lead(2) nitrate was dissolvedin 50.00ml of water a d treated with 25.00 ml of 0.2010M sodium sulfate inoder to determine a content of lead 2 ion. A white precipitate was formed and it was collected and dried. Calculate the amount of precipitate formed in this reaction

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!!!

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