Problem 39 - Chemistry and Chemical Reactivity

Question

Aluminum sulfate ( 1995 worldwide production is about \(\left.3 \times 10^{9} \mathrm{kg}\right)\) is the most commercially important aluminum compound, after aluminum oxide and aluminum hydroxide. Write a balanced equation for the reaction of aluminum oxide with sulfuric acid to give aluminum sulfate. To manufacture \(1.00 \mathrm{kg}\) of aluminum sulfate, what mass (in kilograms) of aluminum oxide and sulfuric acid must be used?

Step-by-Step Solution

Answer

0.298 kg of Al₂O₃ and 0.861 kg of H₂SO₄ are needed.

Step 1: Write the Chemical Equation

First, we need to identify the reactants and products for the chemical reaction. Aluminum oxide (\(\text{Al}_2\text{O}_3\)) reacts with sulfuric acid (\(\text{H}_2\text{SO}_4\)) to produce aluminum sulfate (\(\text{Al}_2(\text{SO}_4)_3\)) and water (\(\text{H}_2\text{O}\)). The unbalanced chemical equation is: \[ \text{Al}_2\text{O}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + \text{H}_2\text{O} \]

Step 2: Balance the Chemical Equation

To balance the chemical equation, ensure the number of each type of atom is the same on both sides of the reaction. The balanced equation is: \[ \text{Al}_2\text{O}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2\text{O} \]

Step 3: Determine Molar Masses

Calculate the molar masses of the reactants and products: \(\text{Al}_2\text{O}_3\): \(102.0 \text{ g/mol}\), \(\text{H}_2\text{SO}_4\): \(98.1 \text{ g/mol}\), and \(\text{Al}_2(\text{SO}_4)_3\): \(342.2 \text{ g/mol}\).

Step 4: Use Stoichiometry to Find Masses

From the balanced equation, 1 mole of \(\text{Al}_2\text{O}_3\) reacts with 3 moles of \(\text{H}_2\text{SO}_4\) to produce 1 mole of \(\text{Al}_2(\text{SO}_4)_3\). To produce 1.00 kg \((1000 \text{ g})\) of \(\text{Al}_2(\text{SO}_4)_3\), the moles needed are \(\frac{1000 \text{ g}}{342.2 \text{ g/mol}} = 2.92 \text{ mol}\).

Step 5: Calculate Required Mass of Reactants

For \(\text{Al}_2\text{O}_3\), use 2.92 moles: \(2.92 \text{ mol} \times 102.0 \text{ g/mol} = 297.84 \text{ g}\). For \(\text{H}_2\text{SO}_4\), use 3 times the moles (since the reaction ratio is 1:3): \(2.92 \text{ mol} \times 3 \times 98.1 \text{ g/mol} = 860.916 \text{ g}\).

Step 6: Convert Grams to Kilograms

Convert the masses from grams to kilograms. \(297.84 \text{ g} = 0.298 \text{ kg}\) of aluminum oxide and \(860.916 \text{ g} = 0.861 \text{ kg}\) of sulfuric acid.

Key Concepts

Balancing Chemical Equations

Balancing chemical equations is a crucial step in stoichiometry, which is essential for using the equations effectively in calculations and experiments. To balance a chemical equation, follow these simple steps:

• Identify all the reactants and products in the reaction. For example, in the reaction of aluminum oxide (\(\text{Al}_2\text{O}_3\)) with sulfuric acid (\(\text{H}_2\text{SO}_4\)), the products are aluminum sulfate (\(\text{Al}_2(\text{SO}_4)_3\)) and water (\(\text{H}_2\text{O}\)
• Write the unbalanced equation, showing all reactants and products. This step helps visualize the changes that occur during the reaction.
• Count the number of atoms for each element on both sides of the equation. Start adjusting the coefficients (numbers in front of compounds) to balance the number of atoms for each element on both sides.
• Use the smallest whole number coefficients to balance the equation. For the aluminum oxide and sulfuric acid reaction, the balanced equation is \[\text{Al}_2\text{O}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2\text{O}\]

Balancing ensures that no atoms are lost or created, adhering to the Law of Conservation of Mass. It's like solving a puzzle; you want to make sure the same number of each type of block is on both sides of the scale.

Molar Mass Calculation

Calculating molar mass is essential to determine the quantities of substances involved in a reaction. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (\(\text{g/mol}\)).

• To calculate the molar mass, sum up the atomic masses of all the atoms in a molecule. For instance, aluminum oxide (\(\text{Al}_2\text{O}_3\)) has two aluminum atoms and three oxygen atoms.
• Using the periodic table, find the atomic mass of aluminum (approximately 27.0 \(\text{g/mol}\)) and oxygen (approximately 16.0 \(\text{g/mol}\)).
• Calculate the molar mass: \(2 \times 27.0 + 3 \times 16.0 = 102.0 \text{ g/mol}\)
• This step is repeated for each substance involved: sulfuric acid (\(\text{H}_2\text{SO}_4\)) and aluminum sulfate (\(\text{Al}_2(\text{SO}_4)_3\)).

Being familiar with these calculations helps you figure out how much of each reactant you need or how much product you can expect to form.

Chemical Reaction Equations

Chemical reaction equations are symbolic representations of chemical reactions showing how reactants transform into products. These equations are not just for noting chemical changes but also serve as the basis for calculating involved substances' masses and moles.A typical chemical equation has:

• Reactants: The starting substances in the reaction, which react with one another.
• An arrow (\(\rightarrow\)): Indicates the conversion of reactants into products.
• Products: The substances formed as a result of the reaction.

For the reaction involving aluminum oxide and sulfuric acid, the equation \[\text{Al}_2\text{O}_3 + 3 \text{H}_2\text{SO}_4 \rightarrow \text{Al}_2(\text{SO}_4)_3 + 3 \text{H}_2\text{O}\]shows how these reactants form new compounds.Understanding chemical equations is vital for:

• Predicting the amounts of products formed.
• Determining the necessary amounts of reactants.
• Studying the energy changes in reactions and ensuring the equations balance to adhere to physical laws like mass and energy conservation.

By mastering the reading and writing of chemical reaction equations, students will develop a strong foundation for exploring more complex chemical processes.