Chapter 4, Problem 4.31PAE

4.29 When Al(OH)₃ reacts with sulfuric acid, the following reaction occurs:

   $\text{2 Al}{\left(\text{OH}\right)}_{\text{3}}+{\text{ 3 H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}+{\text{ 6 H}}_{\text{2}}\text{O}$

If $\text{1}.\text{7 }\times \text{ 1}{0}^{\text{3}}$ g of Al(OH)₃ is combined with 680 g of H₂SO₄, how much aluminum sulfate can form?

Interpretation Introduction

Interpretation:

Mass of aluminium sulfate that can be formed should be determined.

Concept introduction:

• The maximum yield is obtained by reacting all the limiting reactant

Given:

$\begin{array}{l}\text{m=1}\text{.7}\times {\text{10}}^{\text{3}}\text{g}{\text{Al(OH)}}_{\text{3}}\text{=1}\text{.7}\text{kg}{\text{Al(OH)}}_{\text{3}}\\ \text{m=680}\text{g}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=0}\text{.680}\text{kg}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Answer to Problem 4.31PAE

Solution:

$\text{mass}\text{=791}\text{.4 g}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}$.

Explanation of Solution

Use the given amounts to identify the limiting reactant. Then calculate the total amount of possible iron yield.

Step 1: Balance Reaction

Reaction is already balanced

Step 2: Change all mass units to mol units

$\begin{array}{l}\text{Molecular Weight}{\text{Al(OH)}}_{\text{3}}\text{=78}\text{.00 }\text{g/mol}\\ \text{Molecular Weight}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=98}\text{.00}\text{g/mol}\\ {\text{Molecular Weight of Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\text{=342}\text{.15}\text{g/mol}\end{array}$

$\begin{array}{l}\text{mol}{\text{Al(OH)}}_{\text{3}}\text{=}\frac{\text{mass }{\text{Al(OH)}}_{\text{3}}}{\text{Molecular Weight}{\text{Al(OH)}}_{\text{3}}}\text{=}\frac{\text{1}\text{.7}\times {\text{10}}^3}{\text{78}}\text{=}\text{21}\text{.79}\text{mol}\\ \text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\frac{\text{mass}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{Molecular Weight}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{=}\frac{\text{680}}{\text{98}}\text{=}\text{6}\text{.94}\text{mol}\end{array}$

Step 3: Relate stoichiometry

$\text{Ratio=}\frac{\text{mol }{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{mol}{\text{Al(OH)}}_{\text{3}}}\text{=}\frac{\text{3}}{\text{2}}\text{=1}\text{.5}$

Step 4: Actual Ratio

$\text{Ratio=}\frac{\text{mol}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{mol}{\text{Al(OH)}}_{\text{3}}}\text{=}\frac{\text{6}\text{.94}}{\text{21}\text{.79}}\text{=0}\text{.3166}$

Since ratio is smaller than the required, the limiting reactant is sulfuric acid.

Step 4: Calculate Yield

$\begin{array}{l}3\text{mol}\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=1}\text{mol}\text{of}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\\ \text{mol}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\text{=}\left(\frac{\text{1 mol}\text{of}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}}{\text{3 mol }\text{of}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\times \left(6.94\text{mol}{\text{Al(OH)}}_{\text{3}}\right)\\ \text{mol}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\text{=2}\text{.313}\text{mol}\\ \text{mass}\text{=}\text{(mol}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\text{)}\times \text{(MW}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\text{)}\\ \text{mass=(2}\text{.313 }\text{mol)}\times \text{(342}\text{.15 g/mol)}\\ \text{mass=791}\text{.4 g}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}}_{\text{)}}\end{array}$

Conclusion

The maximum amount of product is always equal to the yield of the limiting reactant