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!!!

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."

atomic mass = 39

number of neutrons = 20

number of protons= 19

electronic configuration = 1s2 2s2 2p6 3s2 3p6 4s1

the number of electron in first shell = 2

number of electrons in n=3 shell = 8