# Question 68eca

Jun 19, 2017

$0.00615$ ${\text{mol Pb(NO"_3")}}_{2}$

$0.00615$ ${\text{mol PbCl}}_{2}$ (precipitate)

#### Explanation:

Let's first write the chemical equation for this precipitation reaction:

${\text{Pb(NO"_3")"_2(aq) + 2"KCl" (aq) rarr "PbCl"_2(s) + 2"KNO}}_{3} \left(a q\right)$

To find the moles of each substance present initially, we can use the molarity formula:

$\text{molarity" = "mol solute"/"L soln}$

"mol solute" = ("molarity")("L soln")

• "mol"_ ("Pb(NO"_3")"_2) = (0.806"mol"/(cancel("L")))(0.02150cancel("L soln"))

$= 0.0173$ ${\text{mol Pb(NO"_3")}}_{2}$

• "mol"_ ("KCl") = (0.683"mol"/(cancel("L")))(0.01800cancel("L")) = 0.0123 $\text{mol KCl}$

(I converted the volumes to liters, as is necessary when using this equation.)

To calculate the amount of precipitate that forms, we must first calculate which reactant is limiting. To find this, we simply divide these calculated mole values by the coefficient of the substance in the equation, and whichever number is lower, that reagent is limiting:

• ${\text{Pb(NO"_3")}}_{2}$: (0.0173"mol")/(1 "(coefficient)") = color(red)(0.0173

• $\text{KCl}$: (0.0123"mol")/(2 "(coefficient)") = color(blue)(0.00615

Therefore, $\text{KCl}$ is the limiting reagent, so we'll use the mole values of $\text{KCl}$ for our calculations.

To find the number of moles of ${\text{Pb(NO"_3")}}_{2}$ used, let's use the mole values for $\text{KCl}$ and the coefficients:

0.0123cancel("mol KCl")((1color(white)(l)"mol Pb(NO"_3")"_2)/(2cancel("mol KCl")))

= color(green)(0.00615 color(green)("mol Pb(NO"_3")"_2

Thus, color(green)(0.00615 moles of lead(II) nitrate are used in the reaction.

The moles of precipitate (${\text{PbCl}}_{2}$) expected is calculated the same way:

0.0123cancel("mol KCl")((1color(white)(l)"mol PbCl"_2)/(2cancel("mol KCl")))

= color(purple)(0.00615 color(purple)("mol PbCl"_2#

The amount of moles of precipitate expected is the same amount as lead(II) nitrate used, because their coefficients are both $1$.