Chapter 11, Problem 44A

Interpret the following equation in terms of particles, moles, and mass.

   $\text{4Al}\left(\text{s}\right){\text{+3O}}_{\text{2}}\left(\text{g}\right)\to {\text{2Al}}_{\text{2}}{\text{O}}_{\text{3}}\left(\text{s}\right)$

Interpretation Introduction

Interpretation:

The given reaction in terms of particles, moles and mass needs to be determined.

${\text{4 Al}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

Concept introduction:

The mole concept provides the relation between moles, mass and molar mass. The mathematical expression of mole concept can be written as:

$\text{Mole =}\frac{\text{Mass}}{\text{Molar mass}}$

1 mole of any substance contains $\text{6}\text{.022 }\times {\text{ 10}}^{\text{23}}$ number of particles or atoms or molecules.

Answer to Problem 44A

${\text{4 molecule Al}}_{\text{(s)}}{\text{ + 3molecule O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 molecule Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

${\text{4 particles Al}}_{\text{(s)}}{\text{ + 3particlesO}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 particles Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

$\begin{array}{l}{\text{4 Al}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\\ \text{108g 96g 203}\text{.9g}\end{array}$

Explanation of Solution

According to the given balance chemical reaction:

${\text{4 Al}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

Thus the reaction in terms of molecules:

${\text{4 molecule Al}}_{\text{(s)}}{\text{ + 3molecule O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 molecule Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

Thus the reaction in terms of particles:

${\text{4 particles Al}}_{\text{(s)}}{\text{ + 3particlesO}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 particles Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$

The reaction in terms of mass:

Molar mass of Al = 27.0 g/mol

Molar mass ${\text{O}}_{\text{2(g)}}$ = 32.0 g/mol

Molar mass ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}$ = 101.96 g/mol

$\begin{array}{l}{\text{4 Al}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\\ \text{4 moles Al×}\frac{\text{27}\text{.0g Al }}{\text{1 mole Al}}\text{=108g}\\ {\text{3 moles O}}_{\text{2(g)}}\text{×}\frac{\text{32}{\text{.0g O}}_{\text{2(g)}}}{{\text{1 mole O}}_{\text{2(g)}}}\text{=96g}\\ {\text{2 moles Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\text{×}\frac{\text{101}{\text{.96 g Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}}{{\text{1 mole Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}}\text{=203}\text{.9g}\end{array}$

Conclusion

Thus,

$\begin{array}{l}{\text{4 Al}}_{\text{(s)}}{\text{ + 3 O}}_{\text{2(g)}}\stackrel{}{\to }{\text{ 2 Al}}_{\text{2}}{\text{O}}_{\text{3(s)}}\\ \text{108g 96g 203}\text{.9g}\end{array}$