Chapter 11, Problem 108A

Apply Students conducted a lab to investigate limiting and excess reactants. The students added different volumes of sodium phosphate solution $\left({\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\right)$ to a beaker. They then added a constant volume of cobalt(ll) nitrate solution $\left(\text{CO}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\right)$ stirred the contents, and allowed the beakers to sit overnight. The next day, each beaker had a purple precipitate at the bottom. The students decanted the supernatant from each beaker, divided it into two samples, and added one drop Of sodium phosphate solution to one sample and one drop of cobalt(ll) nitrate solution to the second sample. Their results are shown in Table 11.5.
a. Write a balanced chemical equation for the reaction.
b. Based on the results, identify the limiting reactant and the excess reactant for each trial.

Interpretation Introduction

Interpretation:

The balance chemical equation between sodium phosphate and cobalt (III) nitrate to form purple precipitate needs to be determined.

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactant, which is present in limited amount and determines the amount of product, called as limiting reactant. Whereas another reactant is called as excess reactant.

Answer to Problem 108A

$\text{3 Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + 2 Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ Co}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2(s)}}{\text{ + 6 NaNO}}_{\text{3(aq)}}$

Explanation of Solution

The reaction between sodium phosphate and cobalt (III) nitrate forms purple precipitate. The balance chemical reaction can be written as:

$\text{3 Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + 2 Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ Co}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2(s)}}{\text{ + 6 NaNO}}_{\text{3(aq)}}$

Here the purple precipitate is formed due to cobalt (III) phosphate and sodium nitrate will remain in solution.

Interpretation Introduction

(b)

Interpretation:

The limiting and excess reactant for the reaction between sodium phosphate and cobalt (III) nitrate to form purple precipitate needs to be determined.

$\text{3 Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + 2 Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ Co}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2(s)}}{\text{ + 6 NaNO}}_{\text{3(aq)}}$

Concept introduction:

A chemical reaction involves the conversion of certain molecules (reactant) to new substance (product) by bond breaking and making. One of the reactant, which is present in limited amount and determines the amount of product, called as limiting reactant. Whereas another reactant is called as excess reactant.

Answer to Problem 108A

Trial 1 = ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ must be the limiting reactant whereas $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is excess reactant Trial 1 to 2.

In the trials 2 to 4 = $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is limiting reactant and ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ is inexcess reactant.

Explanation of Solution

The reaction between sodium phosphate and cobalt (III) nitrate forms purple precipitate. The balance chemical reaction can be written as:

$\text{3 Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2(aq)}}{\text{ + 2 Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}\stackrel{}{\to }{\text{ Co}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2(s)}}{\text{ + 6 NaNO}}_{\text{3(aq)}}$

According to trial 1, ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ must be the limiting reactant whereas $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is excess reactant as the ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ because addition of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ does not form precipitate from trial 1 to 2.

In the trials 2 to 4 $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is limiting reactant and ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4(aq)}}$ is in excess because the concentration of $\text{Co}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is constant.