A solution contains 0.133 g of dissolved Lead. How many moles of soduim chloride must be added to the solution to completly precipitate all of the disolved lead? What mass of sodium chloride must be added?

A solved example. Just follow the steps.

http://www.jiskha.com/science/chemistry/stoichiometry.html

a solution contains 0.133 g of dissolved lead. 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?

Ok so first of all equation:

Pb2+ + 2Cl- = PbCl
Then we do some stoich
so we find the mole of the lead n=m/M
n=.133/207.2
n=6.4x10^-4mol
Now by using the mole ratio
for every Lead solution ion there need to be 2 Sodium solution (sodium chloride) ions
therefore double the amount in mol
12.8x10^-4mol
or
1.28x10^-3mol

To determine how many moles of sodium chloride (NaCl) are needed to completely precipitate all of the dissolved lead (Pb), we need to use stoichiometry, specifically the balanced chemical equation for the precipitation reaction.

The balanced chemical equation for the reaction between lead ions and chloride ions is:

Pb²⁺(aq) + 2Cl⁻(aq) → PbCl₂(s)

From the equation, we can see that one mole of lead (Pb) reacts with two moles of chloride ions (Cl⁻) to form one mole of lead chloride (PbCl₂).

First, we need to calculate the number of moles of lead (Pb) present in the solution:

moles of Pb = mass of Pb / molar mass of Pb

Given that the mass of dissolved lead is 0.133 g and the molar mass of lead is approximately 207.2 g/mol:

moles of Pb = 0.133 g / 207.2 g/mol
moles of Pb ≈ 0.000642 mol

Since the molar ratio between lead and chloride ions is 1:2, we need twice the number of moles of chloride ions to precipitate all the lead. Therefore, we need:

moles of Cl⁻ = 2 * moles of Pb
moles of Cl⁻ = 2 * 0.000642 mol
moles of Cl⁻ ≈ 0.00128 mol

From here, we can calculate the mass of sodium chloride (NaCl) needed using the number of moles:

mass of NaCl = moles of NaCl * molar mass of NaCl

The molar mass of sodium chloride is approximately 58.44 g/mol:

mass of NaCl = 0.00128 mol * 58.44 g/mol
mass of NaCl ≈ 0.0748 g

Therefore, approximately 0.0748 grams of sodium chloride must be added to the solution to completely precipitate all of the dissolved lead.