# What mass of precipitate will form if 1.50L of excess Pb(ClO_3)_2 is mixed with 0.100L of 0.270 M NaI? Assume 100% yield.

Jan 4, 2016

$\text{6.22 g}$

#### Explanation:

Start by writing a balanced chemical equation for this double replacement reaction

${\text{Pb"("ClO"_3)_text(2(aq]) + color(red)(2)"NaI"_text((aq]) -> "PbI"_text(2(s]) darr + 2"NaClO}}_{\textrm{3 \left(a q\right]}}$

Lead(II) chlorate will react with sodium iodide to form lead(II) iodide, an insoluble solid that precipitates out of solution, and sodium chlorate.

Notice that you have a $1 : \textcolor{red}{2}$ mole ratio between lead(II) chlorate and sodium iodide. This tells you that the reaction will always consume twice as many moles of sodium iodide than you have moles of lead(II) chlorate that take part in the reaction.

You know that the lead(II) chlorate is in Excess. That means that the sodium iodide will act as a limiting reagent, i.e .it will be *completely consumed8 by the reaction.

The given 100% yield tells you that you can expect all the moles of sodium iodide to actually produce moles of lead(II) iodide.

Use the molarity and volume of the sodium iodide solution to determine how many moles of solute you have

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{N a I} = \text{0.270 M" * "0.100 L" = "0.0270 moles NaI}$

Since you have a $\textcolor{red}{2} : 1$ mole ratio between sodium iodide and lead(II) iodide, the reaction will produce

0.0270 color(red)(cancel(color(black)("moles NaI"))) * "1 mole PbI"_2/(color(red)(2)color(red)(cancel(color(black)("moles NaI")))) = "0.0135 moles PbI"_2

Finally, use lead(II) iodide's molar mass to determine how many grams would contain this many moles

0.0135color(red)(cancel(color(black)("moles PbI"_2))) * "461.01 g"/(1color(red)(cancel(color(black)("mole PbI"_2)))) = color(green)("6.22 g")

The answer is rounded to three sig figs.