Chapter 11.4, Problem 28PP

Interpretation Introduction

Interpretation:

If 14.0g of ${\text{Al(OH)}}_3$ is present in an antacid tablet, the theoretical yield of ${\text{AlCl}}_3$ produced needs to be determined when the tablet reacts with HCl.

Concept introduction:

An actual yield is the mass of a product actually obtained from the reaction. It is usually less than the theoretical yield.

Answer to Problem 28PP

Theoretical yield of ${\text{AlCl}}_3$ is 23.9g.

Explanation of Solution

Given: Aluminium hydroxide is often present in antacids to neutralize the stomach acid (HCl). The reaction occurs as follow:

${\text{Al(OH)}}_3{}_{(s)}+3{\text{HCl}}_{(aq)}\stackrel{}{\to }{\text{AlCl}}_3{}_{(aq)}+3{\text{H}}_2{\text{O}}_{(l)}$

Step 1: Balance the chemical equation:

${\text{Al(OH)}}_3{}_{(s)}+3{\text{HCl}}_{(aq)}\stackrel{}{\to }{\text{AlCl}}_3{}_{(aq)}+3{\text{H}}_2{\text{O}}_{(l)}$

Step 2: Make mass to mole fraction.

For conversion from mass to moles, we divide mass by molecular mass, so in this case moles of ${\text{Al(OH)}}_3$ is:

${\text{14g of Al(OH)}}_3\times \frac{1{\text{ mole Al(OH)}}_3}{78.0{\text{ g Al(OH)}}_3}=0.179{\text{ mole Al(OH)}}_3$

Step 3: Make mole to mole conversion

We know from balanced chemical equation that 1 mole ${\text{Al(OH)}}_3$ produces 1 mole of ${\text{AlCl}}_3$, therefore moles of ${\text{AlCl}}_3$ produced by 0.179 mole of ${\text{Al(OH)}}_3$ is:

$0.179{\text{ mole Al(OH)}}_3\times \frac{1{\text{ mole AlCl}}_3}{\text{1}{\text{.0 mole Al(OH)}}_3}=0.179{\text{ mole AlCl}}_3$

Step 4: Make mole to mass conversion:

From conversion from moles to mass, we multiply moles by molecular mass, so in this case mass of ${\text{AlCl}}_3$ is:

$0.179{\text{ mole AlCl}}_3\times 133.3g{\text{ of AlCl}}_3=23.9g$

Conclusion

Theoretical yield of ${\text{AlCl}}_3$ is 23.9g.