# How many milliliters of 0.10 M ##”Pb”(“NO”_3)_2## are required to react with 75 mL of 0.20 M ##”NaI”##?

##”75 mL”##
##”Pb”(“NO”_ 3)_ (2(aq)) + color(red)(2)”NaI”_ ((aq)) -> 2″NaNO”_ (3(aq)) + “PbI”_ (2(s)) darr##
As you can see, lead(II) nitrate reacts with sodium iodide in a ##1:color(red)(2)## mole ratio, which means that the reaction will always consume twice as many moles of sodium iodide than moles of lead(II) nitrate.
The number of moles of sodium iodide can be determined by using the volume and molarity of the solution
##75 color(red)(cancel(color(black)(“mL”))) * (1 color(red)(cancel(color(black)(“L”))))/(10^3color(red)(cancel(color(black)(“mL”)))) * overbrace(“0.20 moles NaI”/(1color(red)(cancel(color(black)(“L”)))))^(color(blue)(“= 0.20 M”)) = “0.015 moles NaI”##
According to the aforementioned mole ratio, this many moles of sodium iodide will require
##0.015 color(red)(cancel(color(black)(“moles NaI”))) * (“1 mole Pb”(“NO”_3)_2)/(color(red)(2)color(red)(cancel(color(black)(“moles NaI”))))##
##= “0.0075 moles Pb”(“NO”_3)_2##
Now all you have to do is use the molarity of the lead(II) nitrate solution to find the volume that would contain that many moles of solute
##0.0075 color(red)(cancel(color(black)(“moles Pb”(“NO”_3)_2))) * “1 L”/(0.10 color(red)(cancel(color(black)(“mole Pb”(“NO”_3)_2))))##
##= “0.075 L”##
Expressed in milliliters, the answer will be
##color(green)(|bar(ul(color(white)(a/a)color(black)(“volume of Pb”(“NO”_3)_2 = “75 mL”)color(white)(a/a)|)))##
The answer is rounded to two .

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