A solution contains 0.245 of dissolved lead (lead ion or Pb2+.)

How many moles of sodium chloride must be added to the solution to completely precipitate all of the dissolved lead?
What mass of sodium chloride must be added?

Pb^+2 + 2NaCl --> PbCl2 + 2Na^+

Convert 0.245 g Pb^+2 to moles. moles = grams/molar mass

Using the coefficients in the balanced equation, convert moles Pb to moles NaCl. That gives you part a.

Convert moles NaCl to grams. g = moles x molar mass. That gives you part b.

0.00097973

To determine the number of moles of sodium chloride required to completely precipitate all of the dissolved lead (Pb2+), we need to consider the balanced chemical equation of the precipitation reaction between lead ions and sodium chloride.

The balanced equation for the reaction is:
Pb2+ + 2NaCl -> PbCl2 + 2Na+

From the equation, we can see that one mole of lead ions (Pb2+) will react with 2 moles of sodium chloride (NaCl) to form one mole of lead chloride (PbCl2). Therefore, the mole ratio of lead ions to sodium chloride is 1:2.

Given that the solution contains 0.245 moles of lead ions, we can calculate the required moles of sodium chloride using the mole ratio.

Moles of Pb2+ = 0.245 moles
Moles of NaCl = 2 x 0.245 moles = 0.49 moles

Therefore, 0.49 moles of sodium chloride must be added to completely precipitate all of the dissolved lead.

To determine the mass of sodium chloride required, we need to know the molar mass of NaCl, which is 58.44 g/mol.

Mass of NaCl = Moles of NaCl x Molar mass of NaCl
Mass of NaCl = 0.49 moles x 58.44 g/mol ≈ 28.57 g

Therefore, approximately 28.57 grams of sodium chloride must be added to completely precipitate all of the dissolved lead.