# Question d03c9

Jan 31, 2015

$\text{55.4 g}$

#### Explanation:

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.

$2 K I \left(a q\right) + P b {\left(N {O}_{3}\right)}_{2} \left(a q\right) \to P b {I}_{2} \left(s\right) + 2 K N {O}_{3} \left(a q\right)$

This is a double replacement reaction that leads to the formation of an insoluble precipitate, $P b {I}_{2}$.

Notice the mole ratio between $K I$ and $P b {\left(N {O}_{3}\right)}_{2}$: $\text{2 moles}$ of the former must react with $\text{1 mole}$ 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 $K I$) and from moles to grams (for $P b {\left(N {O}_{3}\right)}_{2}$):

55.2 cancel("g") * ("1 mole")/(331.2 cancel("g")) = "0.167 moles" $P b {\left(N {O}_{3}\right)}_{2}$

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 $\text{166 g}$ represents the mass of one mole of potassium iodide. This values comes from potassium iodide's molar mass, which is listed at approximately $\text{166 g /mol}$.

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: