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?
0,00128 moles of Sodium Chloride must be added to the solution to completely precipitate 0,133 g of dissolved Lead.
The chemical reaction for the precipitation of lead from an aqueous solution is:
Pb⁺²(aq) + 2NaCl(aq) → PbCl₂(s) + 2Na⁺(aq)
To calculate how many moles of NaCl are needed, we'll use the following conversion factor to go from grams of Pb⁺² to moles of NaCl using atomic masses and reaction coefficients:
[tex]0,133 g Pb^{+2} * \frac{1 mol Pb^{+2}}{207,2 g Pb^{+2}}* \frac{2 mol NaCl}{1 mol Pb^{+2}}=0,00128 moles NaCl [/tex]
