# Question 0ff49

Sep 19, 2016

Lead(II) nitrate will be the limiting reagent.

#### Explanation:

For starters, you need the balanced chemical equation that describes this double replacement reaction

${\text{Pb"("NO"_ 3)_ (2(aq)) + color(red)(2)"NaI"_ ((aq)) -> "PbI"_ (2(s)) darr + 2"NaNO}}_{3 \left(a q\right)}$

Now, the balanced chemical equation tells you that you need $\textcolor{red}{2}$ moles of sodium iodide for every $1$ mole of lead(II) nitrate that takes part in the reaction.

To calculate the number of moles of lead(II) nitrate present in your $\text{0.150 g}$ sample, use the compound's molar mass

0.150 color(red)(cancel(color(black)("g"))) * ("1 mole Pb"("NO"_ 3)_ 2)/(331.2color(red)(cancel(color(black)("g")))) = "0.000453 moles Pb"("NO"_ 3)_ 2

To calculate the number of moles of sodium iodide, use the volume and the molarity of the solution

225 color(red)(cancel(color(black)("mL"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.100 moles NaI"/(1color(red)(cancel(color(black)("L")))) = "0.0225 moles NaI"

At this point, it should be clear that lead(II) nitrate will act as the limiting reagent because it will be completely consumed before all the moles of sodium iodide can take part in the reaction.

This is the case because $0.0225$ moles of sodium iodide would require

0.0225color(red)(cancel(color(black)("moles NaI"))) * ("1 mole Pb"("NO"_3)_2)/(color(red)(2)color(red)(cancel(color(black)("moles NaI")))) = "0.01125 moles Pb"("NO"_3)_2

Since you have fewer moles of lead(II) nitrate available, lead(II) nitrate will act as the limiting reagent.

The reaction will thus consume $0.000453$ moles of lead(II) nitrate and

0.000453 color(red)(cancel(color(black)("moles Pb"("NO"_3)_2))) * (color(red)(2)color(white)(a)"moles NaI")/(1color(red)(cancel(color(black)("mole Pb"("NO"_3)_2)))) = "0.000906 moles NaI"#

The rest of the sodium iodide will be in excess, i.e. it will not take part in the reaction.