Solution Stoichiometry
To Begin
1. 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
»1L
»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