Question #d03c9

Every time you have to solve problems in which you are asked to determine how much of something is needed to react with a given mass of something else, you must think moles and, more specifically, mole ratios.

Start with the balanced chemical equation

#2KI(aq) + Pb(NO_3)_2(aq) -> PbI_2(s) + 2KNO_3(aq)#

This is a double replacement reaction that leads to the formation of an insoluble precipitate, #PbI_2#.

Notice the mole ratio between #KI# and #Pb(NO_3)_2#: #"2 moles"# of the former must react with #"1 mole"# of the latter in order for the reaction to take place.

Use the molar masses of the compounds to go from grams to moles (for #KI#) and from moles to grams (for #Pb(NO_3)_2#):

#55.2 cancel("g") * ("1 mole")/(331.2 cancel("g")) = "0.167 moles"# #Pb(NO_3)_2#

This means that you need

#0.167 cancel("moles"color(white)(a) Pb(NO_3)_2) * ("2 moles " KI)/(1 cancel("mole"color(white)(a) Pb(NO_3)_2)) = "0.334 moles"color(white)(a) KI#

The mass will be

#0.334 cancel("moles") * ("166 g")/(1 cancel("mole")) = "55.4 g"color(white)(a) KI#

Here #"166 g"# represents the mass of one mole of potassium iodide. This values comes from potassium iodide's molar mass, which is listed at approximately #"166 g /mol"#.

As a conclusion, always go to moles, it'll make your life easier, and remember to use mole ratios.

Here's a video of the reaction, if you're curios about what the precipitate looks like: