Chapter 4, Problem 91QP

Interpretation Introduction

Interpretation:

The pair of combination that produces a greater mass of the solid product is to be determined.

Concept introduction:

In a precipitation reaction, two ionic compounds in a solution react to form an insoluble compound, which is called the precipitate.

The formation of the precipitate can be predicted by using solubility rules.

The mass of the precipitate can be determined from the stoichiometry of the reaction.

Answer to Problem 91QP

Solution:

(a) Combination 1

(b) Combination 2

(c) Combination 2

Explanation of Solution

a)

Combination 1: $105.5\text{ mL 1}\text{.508 M Pb}{\left({\text{NO}}_3\right)}_2\text{and 250}\text{.0 mL 1}\text{.2075 M KCl}$

Combination 2: $\text{138}\text{.5 mL 1}\text{.469 M Pb}{\left({\text{NO}}_3\right)}_2\text{and 100}\text{.0 mL 2}\text{.115 M KCl}$

In Combination 1:

The precipitation reaction between lead nitrate and potassium chloride is as follows:

$\text{2KCl}\left(aq\right)+\text{Pb}{\left({\text{NO}}_3\right)}_2\left(aq\right)\to {\text{PbCl}}_2\left(s\right)+2{\text{KNO}}_3\left(aq\right)$

From the above balanced equation, one mole of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ reacts with two moles of KCl.

First, calculate the number of moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ using the given volume and concentration:

$\begin{array}{c}\text{Moles of Pb}{\left({\text{NO}}_3\right)}_2\text{= 1}\text{.508 M}\times \text{0}\text{.1055 L}\\ =0.159094\text{ mol}\end{array}$

Now, calculate the number of moles of KCl.

$\begin{array}{c}\text{Moles of KCl = 1}\text{.2075 M}\times \text{0}\text{.250 L}\\ =0.301875\text{ mol}\end{array}$

Hence, KCl is a limiting reagent that determines the amount of ${\text{PbCl}}_2$ formed.

$\begin{array}{c}{\text{Moles of PbCl}}_2\text{ = mol KCl}\times \frac{{\text{1 mol PbCl}}_{\text{2}}}{\text{2 mol KCl}}\\ =0.301875\text{ mol}\times \frac{{\text{1 mol PbCl}}_{\text{2}}}{\text{2 mol KCl}}\\ =0.150938\text{ mol}\end{array}$

Convert moles to grams by multiplying the moles with the molar mass of ${\text{PbCl}}_2$

$\begin{array}{c}{\text{Mass of PbCl}}_2\text{ = 0}\text{.150938 mol}\times \text{278}{\text{.1 g/mol PbCl}}_2\\ \text{= 41}\text{.98 g}\end{array}$

Hence, the mass of the ${\text{PbCl}}_2$ formed is $\text{41}\text{.98 g}$.

In Combination 2,

Calculate the number of moles of $\text{Pb}{\left({\text{NO}}_3\right)}_2$ using the given volume and concentration

$\begin{array}{c}\text{Moles of Pb}{\left({\text{NO}}_3\right)}_2\text{ = 1}\text{.469 M}\times \text{0}\text{.1385 L}\\ =0.203457\text{ mol}\end{array}$

Now, calculate the number of moles of KCl.

$\begin{array}{c}\text{Moles of KCl = 2}\text{.115 M}\times \text{0}\text{.100 L}\\ =0.2115\text{ mol}\end{array}$

KCl is a limiting reagent that determines the amount of ${\text{PbCl}}_2$ formed.

$\begin{array}{c}{\text{Moles of PbCl}}_2\text{ = mol KCl}\times \frac{{\text{1 mol PbCl}}_{\text{2}}}{\text{2 mol KCl}}\\ =0.2115\text{ mol}\times \frac{{\text{1 mol PbCl}}_{\text{2}}}{\text{2 mol KCl}}\\ =0.10575\text{ mol}\end{array}$

Now convert the number of moles to grams by multiplying the number of moles with the molar mass of PbCl₂

$\begin{array}{c}{\text{g PbCl}}_2\text{= 0}\text{.10575 mol×278}{\text{.1 g/mol PbCl}}_2\\ \text{= 29}\text{.41 g}\end{array}$

Hence, the mass of the PbCl₂ formed is $\text{29}\text{.41 g}$.

Therefore, combination 1 will produce a greater mass of the product.

b)

Combination 1: $\text{32}\text{.25 mL 0}\text{.9475}M{\text{ Na}}_3{\text{PO}}_4\text{ and 92}\text{.75 mL 0}\text{.7750}M\text{ Ca}{\left({\text{NO}}_3\right)}_2$

Combination 2: $\text{52}\text{.50 mL 0}\text{.6810}M{\text{ Na}}_3{\text{PO}}_4\text{ and 39}\text{.50 mL 1}\text{.555}M\text{ Ca}{\left({\text{NO}}_3\right)}_2$

Combination 1:

The precipitation reaction between sodium phosphate and calcium nitrate is as follows:

${\text{2Na}}_3{\text{PO}}_4\left(aq\right)+3\text{Ca}{\left({\text{NO}}_3\right)}_2\left(aq\right)\to {\text{Ca}}_3{\left({\text{PO}}_4\right)}_2\left(s\right)+6{\text{NaNO}}_3\left(aq\right)$

From the above balanced equation, two moles of ${\text{Na}}_3{\text{PO}}_4$ react with three moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$.

First, calculate the number of moles of Na₃PO₄ using the given volume and concentration:

$\begin{array}{c}{\text{Moles of Na}}_3{\text{PO}}_4\text{= 0}\text{.9475 M}\times \text{0}\text{.03225 L}\\ =0.030557\text{ mol}\end{array}$

Now, calculate the number of moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$:

$\begin{array}{c}\text{Moles of Ca}{\left({\text{NO}}_3\right)}_2\text{= 0}\text{.7750 M}\times \text{0}\text{.09275 L}\\ =0.071881\text{ mol}\end{array}$

Here, Na₃PO₄ is a limiting reagent that determines the amount of product formed.

$\begin{array}{c}{\text{Moles of Ca}}_3{\left({\text{PO}}_4\right)}_2{\text{ = mol Na}}_3{\text{PO}}_4\times \frac{{\text{1 mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{{\text{2 mol Na}}_3{\text{PO}}_4}\\ =0.030557\text{ mol}\times \frac{{\text{1 mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{{\text{2 mol Na}}_3{\text{PO}}_4}\\ =0.015279\text{ mol}\end{array}$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$:

$\begin{array}{c}{\text{Mass of Ca}}_3{\left({\text{PO}}_4\right)}_2\text{ = 0}\text{.015279 mol × 310}{\text{.18 g/mol Ca}}_3{\left({\text{PO}}_4\right)}_2\\ \text{= 4}\text{.74 g}\end{array}$

Hence, the mass of the ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ formed is $4.74\text{ g}$.

Combination 2:

Calculate the number of moles of ${\text{Na}}_3{\text{PO}}_4$ using the given volume and concentration:

$\begin{array}{c}{\text{Moles of Na}}_3{\text{PO}}_4\text{ = 0}\text{.6810 M }\times \text{ 0}\text{.05250 L}\\ =0.0357525\text{ mol}\end{array}$

Now, calculate the number of moles of $\text{Ca}{\left({\text{NO}}_3\right)}_2$:

$\begin{array}{c}\text{Moles of Ca}{\left({\text{NO}}_3\right)}_2\text{= 0}\text{.7750 M}\times \text{0}\text{.09275 L}\\ =0.0614225\text{ mol}\end{array}$

Here, ${\text{Na}}_3{\text{PO}}_4$ is a limiting reagent that determines the amount of product formed.

$\begin{array}{c}{\text{Moles of Ca}}_3{\left({\text{PO}}_4\right)}_2{\text{ = mol Na}}_3{\text{PO}}_4\times \frac{{\text{1 mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{{\text{2 mol Na}}_3{\text{PO}}_4}\\ =0.0357525\text{ mol}\times \frac{{\text{1 mol Ca}}_3{\left({\text{PO}}_4\right)}_2}{{\text{2 mol Na}}_3{\text{PO}}_4}\\ =0.01787625\text{ mol}\end{array}$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$:

$\begin{array}{c}{\text{g Ca}}_3{\left({\text{PO}}_4\right)}_2\text{= 0}\text{.01787625 mol × 310}{\text{.18 g/mol Ca}}_3{\left({\text{PO}}_4\right)}_2\\ \text{= 5}\text{.54 g}\end{array}$

Hence, the mass of the ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$ formed is $5.54\text{ g}$.

Therefore, combination 2 produces a greater mass of the product.

c)

Combination 1: $\text{29}\text{.75 mL 1}{\text{.575 M AgNO}}_3\text{ and 25}\text{.00 mL 2}{\text{.010 M BaCl}}_2$

Combination 2: $\text{52}\text{.80 mL 2}{\text{.010 M AgNO}}_3\text{ and 73}\text{.50 mL 0}{\text{.7500 M BaCl}}_2$

Combination 1:

The precipitation reaction between silver nitrate and barium chloride is as follows:

${\text{BaCl}}_2\left(aq\right)+2{\text{AgNO}}_3\left(aq\right)\to 2\text{AgCl}\left(s\right)+\text{Ba}{\left({\text{NO}}_3\right)}_2\left(aq\right)$

From the above balanced equation, one mole of BaCl2 reacts with two moles of ${\text{AgNO}}_3$.

First, calculate the number of moles of ${\text{BaCl}}_2$ using the given volume and concentration:

$\begin{array}{c}{\text{Moles of BaCl}}_2\text{ = 2}\text{.010 M}\times \text{0}\text{.0250 L}\\ =0.05025\text{ mol}\end{array}$

Now, calculate the number of moles of ${\text{AgNO}}_3$.

$\begin{array}{c}{\text{Moles of AgNO}}_3\text{ = 1}\text{.575 M}\times \text{0}\text{.02975 L}\\ =0.0468563\text{ mol}\end{array}$

Hence, ${\text{AgNO}}_3$ is a limiting reagent that determines the amount of AgCl formed.

$\begin{array}{c}{\text{Moles of AgCl = mol AgNO}}_3\times \frac{\text{2 mol AgCl}}{{\text{2 mol AgNO}}_3}\\ =0.0468563{\text{ mol AgNO}}_3\times \frac{\text{2 mol AgCl}}{{\text{2 mol AgNO}}_3}\\ =0.0468563\text{ mol}\end{array}$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$\begin{array}{c}\text{g AgCl = 0}\text{.0468563 mol × 143}\text{.32 g/mol AgCl }\\ \text{= 6}\text{.72 g}\end{array}$

Hence, the mass of the AgCl formed is $\text{6}\text{.72 g}$.

Combination 2:

Calculate the number of moles of BaCl₂ using the given volume and concentration:

$\begin{array}{c}{\text{Moles of BaCl}}_2\text{ = 0}\text{.7500 M}\times \text{0}\text{.07350 L}\\ =0.055125\text{ mol}\end{array}$

Now, calculate the number of moles of AgNO₃.

$\begin{array}{c}{\text{Moles of AgNO}}_3\text{ = 2}\text{.010 M}\times \text{0}\text{.05280 L}\\ =0.106128\text{ mol}\end{array}$

Hence, ${\text{AgNO}}_3$ is a limiting reagent that determines the amount of AgCl formed.

$\begin{array}{c}{\text{Moles of AgCl = mol AgNO}}_3\times \frac{\text{2 mol AgCl}}{{\text{2 mol AgNO}}_3}\\ =0.106128{\text{ mol AgNO}}_3\times \frac{\text{2 mol AgCl}}{{\text{2 mol AgNO}}_3}\\ =0.106128\text{ mol}\end{array}$

Convert the number of moles to grams by multiplying the number of moles with the molar mass of AgCl:

$\begin{array}{c}\text{g AgCl = 0}\text{.106128 mol×143}\text{.32 g/mol AgCl }\\ \text{= 15}\text{.21 g}\end{array}$

Hence, the mass of the AgCl formed is $\text{15}\text{.21 g}$.

Therefore, combination 2 produces a greater mass of the product.