Chapter 4, Problem 4.24PAE

4.24 Ammonia gas can be prepared by the reaction

   $\text{CaO}\left(s\right)+{\text{ 2 NH}}_{\text{4}}\text{Cl}\left(s\right)\to {\text{ 2 NH}}_{\text{3}}\left(g\right)+{\text{ H}}_{\text{2}}\text{O}\left(g\right)+{\text{ CaCl}}_{\text{2}}\left(s\right)$

If 112 g of CaO reacts with 224 g of NH₄Cl, how many moles of reactants and products are there when the reaction is complete?

Interpretation Introduction

Interpretation:

To calculate how many moles of reactants and products will be left after the completion of reaction.

Concept introduction:

• Reactants decrease in moles, Products increase in moles.
• Stoichiometry relates all species to their atomic ratios.
• Limiting reactant will never be leftover if 100 % completion.
• Excess reactant will always be leftover.

Given:

$\begin{array}{l}\text{m }=112gofCaO\\ \text{m }=224gofN{H}_{4}Cl\end{array}$

Answer to Problem 4.24PAE

Solution:

$\begin{array}{l}\mathrm{Re}actatns:\\ 0.20molN{H}_{4}Cl\\ \mathrm{Pr}oducts:\\ 4.0molN{H}_3\\ 2.0mol{H}_{2}O\\ 2.0molCaCl\end{array}$

Explanation of Solution

When a reaction occurs, there are reactants reacting in order to form new products. The reactants are not typically added in a 100 % stoichiometric ratio, therefore, there is always a limiting and an excess reactant.

It is much easier to convert to molar units, rather than mass units, in order to verify the reactants and products present.

Step 1: Balance Reaction

The reaction is already balanced

Step 2: Change all mass units to mol units

$MolecularWeight(MW)CaO=56.0774g/mol$

$MolecularWeight(MW)ofN{H}_{4}Cl=53.491g/mol$

$\begin{array}{l}MolesofCaO=\frac{massofCaO}{MWofCaO}=\frac{112g}{56.0774g/mol}=2.0mol\hfill \\ MolesofN{H}_{4}Cl=\frac{massofN{H}_{4}Cl}{MWofN{H}_{4}Cl}=\frac{224g}{53.491g/mol}=4.20mol\hfill \end{array}$

Step 3: Relate stoichiometry

$1molofCaO:2molofN{H}_{4}Cl:2molofN{H}_3:1molof{H}_{2}O:1molofCaC{l}_2$

Step 4: Moles reacted

Identify the limiting reactant ( $CaO$

$molCaOleft=2.0-2.0=0.0$

Excess reactant of $N{H}_{4}Cl$

$molN{H}_{4}Cl=4.20-\left(\frac{2molN{H}_{4}Cl}{1molCaO}\right)\left(2.0molCaO\right)=0.20molN{H}_{4}Cl$

Step 5: Moles produced

Moles of product $N{H}_3$ formed

$molN{H}_3=0+\left(\frac{2molN{H}_3}{1molCaO}\right)\left(2.0molCaO\right)=4.0molN{H}_3$

Moles of product $H_{2}O$ H2O formed

$mol{H}_{2}O=0+\left(\frac{1mol{H}_{2}O}{1molCaO}\right)\left(2.0molCaO\right)=2.0mol{H}_{2}O$

Moles of product CaCl2 formed

$molCaCl=0+\left(\frac{1molCaCl}{1molCaO}\right)\left(2.0molCaO\right)=2.0molCaCl$

Conclusion

Use stoichiometry to verify the ratio between species, that is, reactants and products after the reaction is over.