Solution Stoichiometry

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Presentation transcript:

Solution Stoichiometry

To Begin When aqueous solutions of Sodium Sulfate and Lead (II) Nitrate are mixed, Lead Sulfate precipitates. Calculate the mass of Lead Sulfate formed when 1.25L of .05M Lead (II) Nitrate and 2.0 L of .025M Sodium Sulfate are mixed. Write and balance the equation for the reaction. It may be easiest to write the Ionic equation first. Na1+SO42-+Pb2+NO31-↔Pb2+SO42-+Na1+ + NO31-

Na1+SO42-+Pb2+NO31-↔Pb2+SO42-+Na1+ + NO31- Now that I know the charge of each element, I can write the molecular equation. Na2SO4+Pb(NO3)2↔PbSO4+2Na+2NO3 Why not combine these two? We don’t need to. We know that we have a precipitate of Lead Sulfate, these other things are just Spectator Ions and they are not reacting.

SPECTATOR IONS?? Spectator ions are not participating in the reaction. If I look at my solubility rules, these are the things that are soluble. They are still in the solution watching the rest of the reaction.

Na2SO4+Pb(NO3)2↔PbSO4+2Na+2NO3 Now I can write the balanced Ionic Equation. Because these are all ionic compounds, when I put them into water they will break apart. 2Na1+SO42-+Pb2+2NO31-↔PbSO4+2Na1+ + 2NO31- So I basically just moved the subscripts to the place of the coefficients to show how many of each individual I have in the equation. I kept Lead (II) Sulfate together because it precipitated.

Now for the Net Ionic Equation Now I am only interested in what actually reacted in the equation. Any spectator I will leave out. SO42-+Pb2+↔PbSO4

MOLES Now I need to know how many moles of each ion can be formed. 1.25L * .05 moles = .0625 moles of Pb 1L 2L * .025 moles= .05 moles SO4 I can really only make .05 moles of PbSO4

Mole Ratio When I look at the molecular equation for this reaction ,I notice that the reactants and my product all have a 1:1 ratio so I can go straight to grams. .05 mol PbSO4 *303.3 g/mol =15.2gPbSO4