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

2 Answers
Apr 19, 2017

Answer:

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.

Apr 19, 2017

Answer:

#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#