Question

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

Answer 1

I think you mean to ask how much `KI` you need to precipitate the `Pb^(2+)` out from aqueous solution..............

Explanation:

`Pb(NO_3)_2(aq) + 2KI(aq) rarr PbI_2(s)darr + 2KNO_3(aq)`

`"Lead iodide"` precipitates from aqueous solution as a bright yellow powder. And it is pretty insoluble stuff.

With respect to `Pb(NO_3)_2`, we have a molar quantity of `(0.450*g)/(331.20*g*mol^-1)=1.36xx10^-3*mol` .

And thus, for metathesis, we need a TWICE molar quantity of `"potassium iodide"`, i.e. `2xx1.36xx10^-3*molxx166.0*g*mol^-1=0.451*g` `KI`.

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

Answer 2

`0.451" g of "KI`

Explanation:

Given: `0.450" g "Pb(NO_3)_2`

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

`(0.450" g "Pb(NO_3)_2)/1`

We look up the molar mass of `Pb(NO_3)_2` and add it as a factor of moles per gram to the conversion:

`(0.450" g "Pb(NO_3)_2)/1(1" mole "Pb(NO_3)_2)/(331.2" g "Pb(NO_3)_2)`

Please observe how the units cancel:

`(0.450cancel(" g "Pb(NO_3)_2))/1(1" mole "Pb(NO_3)_2)/(331.2cancel(" g "Pb(NO_3)_2))`

From the balanced equation, we observe that 2 moles of `KI` is needed for 1 mole of `Pb(NO_3)_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:

`(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))`

Finally, we look up the molar mass of `KI` 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:

`(0.450cancel(" 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 "KI))`

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`