# Are there examples of grams of precipitate problems? I am not sure what to call them other than that.

Mar 12, 2015

Yes. Precipitation reactions are a type of double replacement (displacement) reaction in which the pattern of reaction is AX + BY $\rightarrow$ AY + BX. A and B are cations, and X and Y are anions. The cations and anions switch partners. These reactions have a product that is a precipitate, an insoluble gas, or water.

You are asking about a double replacement stoichiometry problem involving a precipitate. The following is an example.

Lead(II) nitrate and potassium iodide, both of which are in aqueous solution, react to produce solid lead(II) iodide and aqueous potassium iodide. If 2.89 grams of lead (II) nitrate and 1.05 grams of potassium iodide react, how many grams of the lead(II) iodide precipitate are produced?

Step 1
Write a balanced chemical equation for this reaction.
"Pb(NO"_3)_2("aq") + $\text{2KI(aq)}$ $\rightarrow$ "PbI"_2("s") + "2KNO"_3("aq")

Step 2
Convert given grams of each reactant to moles by dividing grams by molar mass.
Molar mass of "Pb(NO"_3)_2 = $\text{331.2 g/mol}$
Molar mass of $\text{KI}$ = $\text{166.0 g/mol}$

"2.89g Pb(NO"_3)_2$\div$"331.2g/mol Pb(NO"_3)_2 = "0.008726 mol Pb(NO"_3)_2

$\text{1.05g KI}$$\div$$\text{166.0g/mol KI}$ = $\text{0.006325 mol KI}$

Step 3
Determine how many moles of $\text{PbI"_2}$ each reactant can produce using the number of moles of each reactant and the mole ratios from the balanced equation.
Mole ratio for "Pb(NO"_3)_2 and $\text{PbI"_2}$ = $\text{1 mol Pb(NO"_3)_2:1 "mol PbI"_2}$.
Mole ratio for $\text{KI}$ and $\text{PbI"_2}$ = $\text{2 mol KI:1 mol PbI"_2}$

Multiply the known moles of each compound times the mole ratio that has the ${\text{PbI}}_{2}$ on top.

"0.008726 mol Pb(NO"_3)_2 x $\text{1 mol PbI2"/"1 mol Pb(NO3)2}$ = ${\text{0.008726 mol PbI}}_{2}$

$\text{0.006325 mol KI}$ x $\text{1 mol PbI2"/"2 mol KI}$ = ${\text{0.0031625 mol PbI}}_{2}$

Step 4
$\text{KI}$ is the limiting reactant since it produces the fewest moles of ${\text{PbI}}_{2}$, therefore the greatest number of moles of ${\text{PbI}}_{2}$ that can be produced in this reaction is ${\text{0.0031625 mol PbI}}_{2}$.

Now determine the mass in grams of ${\text{PbI}}_{2}$ that can be produced by multiplying the moles of ${\text{PbI}}_{2}$ times its molar mass, which is $\text{461.0 g/mol}$.

${\text{0.0031625 mol PbI}}_{2}$ x $\text{461.0 g PbI2"/"1 mol PbI2}$ = ${\text{1.46 g PbI}}_{2}$

Note: Molar masses were obtained from Wikipedia.