# Given a 0.450*g mass of lead nitrate, what mass of sodium iodide is required to precipitate the metal as it iodide salt?

Apr 19, 2017

I think you mean to ask how much $K I$ you need to precipitate the $P {b}^{2 +}$ out from aqueous solution..............

#### Explanation:

$P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) + 2 K I \left(a q\right) \rightarrow P b {I}_{2} \left(s\right) \downarrow + 2 K N {O}_{3} \left(a q\right)$

$\text{Lead iodide}$ precipitates from aqueous solution as a bright yellow powder. And it is pretty insoluble stuff.

With respect to $P b {\left(N {O}_{3}\right)}_{2}$, we have a molar quantity of $\frac{0.450 \cdot g}{331.20 \cdot g \cdot m o {l}^{-} 1} = 1.36 \times {10}^{-} 3 \cdot m o l$ .

And thus, for metathesis, we need a TWICE molar quantity of $\text{potassium iodide}$, i.e. $2 \times 1.36 \times {10}^{-} 3 \cdot m o l \times 166.0 \cdot g \cdot m o {l}^{-} 1 = 0.451 \cdot g$ $K I$.

Anyway, if I have misinterpreted your problem, send me a beatdown.

Apr 19, 2017

$0.451 \text{ g of } K I$

#### Explanation:

Given: $0.450 \text{ g } P b {\left(N {O}_{3}\right)}_{2}$

We begin the factor-label method by writing the given over 1:

$\frac{0.450 \text{ g } P b {\left(N {O}_{3}\right)}_{2}}{1}$

We look up the molar mass of $P b {\left(N {O}_{3}\right)}_{2}$ and add it as a factor of moles per gram to the conversion:

$\left(0.450 \text{ g "Pb(NO_3)_2)/1(1" mole "Pb(NO_3)_2)/(331.2" g } P b {\left(N {O}_{3}\right)}_{2}\right)$

Please observe how the units cancel:

$\left(0.450 \cancel{\text{ g "Pb(NO_3)_2))/1(1" mole "Pb(NO_3)_2)/(331.2cancel(" g } P b {\left(N {O}_{3}\right)}_{2}}\right)$

From the balanced equation, we observe that 2 moles of $K I$ is needed for 1 mole of $P b {\left(N {O}_{3}\right)}_{2}$, therefore, we add this as a factor to the conversion:

(0.450cancel(" g "Pb(NO_3)_2))/1(1" mole "Pb(NO_3)_2)/(331.2cancel(" g "Pb(NO_3)_2))(2" moles "KI)/(1" mole "Pb(NO_3)_2)

Again, please observe how we have cancelled the units as we progress toward the conversion:

$\left(0.450 \cancel{\text{ g "Pb(NO_3)_2))/1(cancel(1" mole "Pb(NO_3)_2))/(331.2cancel(" g "Pb(NO_3)_2))(2" moles "KI)/(cancel(1" mole } P b {\left(N {O}_{3}\right)}_{2}}\right)$

Finally, we look up the molar mass of $K I$ and add it as a factor of grams per mole to the conversion:

(0.450cancel(" g "Pb(NO_3)_2))/1(cancel(1" mole "Pb(NO_3)_2))/(331.2cancel(" g "Pb(NO_3)_2))(2" moles "KI)/(cancel(1" mole "Pb(NO_3)_2))(166.0" g "KI)/(1" mole "KI)

With a final cancellation of units, please observe that the only units that remain are grams of Potassium Iodide:

$\left(0.450 \cancel{\text{ g "Pb(NO_3)_2))/1(cancel(1" mole "Pb(NO_3)_2))/(331.2cancel(" g "Pb(NO_3)_2))(2cancel(" moles "KI))/(cancel(1" mole "Pb(NO_3)_2))(166.0" g "KI)/(1cancel(" mole } K I}\right)$

We merely perform the multiplications and divisions to obtain the answer:

0.450/1(1/331.2)(2/1)(166.0" g "KI)/1= 0.451" g "KI