A solution contains 0.142 g of dissolved lead. How many moles of sodium chloride must be added to the solution to completely precipitate all of the dissolved lead?
balance the equation:
2NaCl(aq) + Pb >>>PbCl2 + Na+(aq)
so two moles of sodium chloride for each mole of Pb.
how many moles of Pb do you have?
Pb^2+(aq) + 2Cl^-*aq) ==> PbCl2(s)
mols Pb^2+ = grams/atomic mass = ?
mols NaCl must be twice mols Pb.
To determine the number of moles of sodium chloride needed to completely precipitate all of the dissolved lead, we need to consider the stoichiometry of the reaction.
The balanced chemical equation for the reaction between lead and sodium chloride is:
Pb(NO3)2 + 2NaCl -> PbCl2 + 2NaNO3
From the equation, we can see that for every 1 mole of lead (Pb), we need 2 moles of sodium chloride (NaCl) to completely precipitate the lead as lead chloride (PbCl2).
First, we need to convert the mass of lead (0.142 g) to moles. To do this, we divide the mass of lead (in grams) by the molar mass of lead (207.2 g/mol):
moles of lead = 0.142 g / 207.2 g/mol = 0.0006867 mol
Next, we determine the number of moles of sodium chloride needed by multiplying the number of moles of lead by the stoichiometric ratio between lead and sodium chloride:
moles of sodium chloride = moles of lead * (2 moles NaCl / 1 mole Pb) = 0.0006867 mol * 2 = 0.0013734 mol
Therefore, to completely precipitate all of the dissolved lead, we need 0.0013734 moles of sodium chloride.