What volume of .150 M potassium iodide solution will completely react with .155 L of .112 M lead (II) nitrate? equation: 2KI (aq) + Pb(NO3)2 (aq) ----> 2KNO3 (aq) + PbI2 (s)

(I'll use the balanced equation given in the question for this answer).

To solve this equation, let's first use the molarity equation to solve for the moles of lead(II) nitrate:

#"moles solute" = ("molarity")("liters solution")#

#"mol Pb(NO"_3")"_2 = (0.112"mol"/(cancel("L")))(0.155cancel("L"))#

#= color(red)(0.0174# #color(red)("mol Pb(NO"_3")"_2#

Now, we can use the coefficients of the chemical equation to calculate the relative number of moles of potassium iodide, #"KI"#:

#color(red)(0.0174)cancel(color(red)("mol Pb(NO"_3")"_2))((2color(white)(l)"mol KI")/(1cancel("mol Pb(NO"_3")"_2))) = color(blue)(0.0347# #color(blue)("mol KI"#

Now, let's use the molarity equation again to solve for the liters of the potassium iodide solution:

#"liters solution" = ("moles solute")/("molarity")#

#"L KI solution" = (0.0347cancel("mol KI"))/(0.150(cancel("mol"))/("L")) = color(purple)(0.231# #color(purple)("L KI solution"#

Therefore, you would need #color(purple)(0.231# #sfcolor(purple)("liters"# of the #0.150M# potassium iodide solution to completely react with the given lead(II) nitrate solution.