# Question 5a225

Jul 6, 2017

$4.13$ $\text{g excess KCl}$

#### Explanation:

We know that $4.6$ ${\text{g PbCl}}_{2}$ were recovered, and that $\text{KCl}$ is the excess reactant (according to the image).

We can convert this value to moles of ${\text{PbCl}}_{2}$ using its molar mass ($278.1$ $\text{g/mol}$):

4.6cancel("g PbCl"_2)((1color(white)(l)"mol PbCl"_2)/(278.1cancel("g PbCl"_2))) = color(red)(0.0165 color(red)("mol PbCl"_2

We can then use the coefficients of the equation to determine the moles of $\text{KCl}$ used up:

color(red)(0.0165)cancel(color(red)("mol PbCl"_2))((2color(white)(l)"mol KCl")/(1cancel("mol PbCl"_2))) = color(green)(0.0331 color(green)("mol KCl"

Lastly (using molar mass of $\text{KCl}$), let's convert this value to grams, and subtract from the original value to determine how much $\text{KCl}$ was not used:

color(green)(0.0331)cancel(color(green)("mol KCl"))((74.55color(white)(l)"g KCl")/(1cancel("mol KCl"))) = color(purple)(2.47 color(purple)("g KCl"

Subtracting from original value, we have

$6.60$ $\text{g KCl}$ - color(purple)(2.47 color(purple)("g KCl" = color(blue)(4.13 color(blue)("g excess KCl"

which, rounded to one decimal place like you have been doing, is color(blue)(4.1 color(blue)("g KCl"#.