Question

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from \(14.2 \mathrm{~g}\) of aluminum sulfide?

Short answer

(a) The balanced chemical equation for the reaction between aluminum sulfide and water is: Al2S3 + 6H2O → 2Al(OH)3 + 3H2S (b) Approximately 14.76 grams of aluminum hydroxide are obtained when 14.2 grams of aluminum sulfide reacts with water.

Step-by-step solution

Write the Unbalanced Chemical Equation

The reaction between aluminum sulfide (Al2S3) and water (H2O) produces aluminum hydroxide (Al(OH)3) and hydrogen sulfide (H2S). The unbalanced chemical equation is: Al2S3 + H2O → Al(OH)3 + H2S

Balance the Chemical Equation

To balance the equation, we'll adjust the coefficients so that the number of atoms of each element is the same on both sides. Al2S3 + 6H2O → 2Al(OH)3 + 3H2S Now the equation is balanced: there are 2 aluminum, 6 oxygen, 6 hydrogen, and 3 sulfur atoms on both sides of the equation.

Find Moles of Aluminum Sulfide Given

To determine the number of moles of aluminum sulfide given, we'll use the mass of aluminum sulfide and its molar mass. The molar mass of Al2S3 is: (2 x 26.98 g/mol for Al) + (3 x 32.06 g/mol for S) = 150.16 g/mol We are given 14.2 grams of aluminum sulfide, so we can find the moles as follows: \( \text{moles of Al2S3} = \frac{\text{mass of Al2S3}}{\text{molar mass of Al2S3}} \) \( \text{moles of Al2S3} = \frac{14.2 \mathrm{~g}}{150.16 \mathrm{~g/mol}} \approx 0.0946 \mathrm{~mol} \)

Calculate Moles of Aluminum Hydroxide Formed

Using the balanced chemical equation, we can find the moles of aluminum hydroxide formed through stoichiometry. For each mole of aluminum sulfide, two moles of aluminum hydroxide are formed: \( \text{moles of Al(OH)3} = 2 \times \text{moles of Al2S3} \) \( \text{moles of Al(OH)3} = 2 \times 0.0946 \mathrm{~mol} \approx 0.1892 \mathrm{~mol} \)

Calculate Mass of Aluminum Hydroxide Formed

Finally, we will determine the mass of aluminum hydroxide formed using the number of moles we found in the previous step and the molar mass of Al(OH)3 which is: (1 x 26.98 g/mol for Al) + (3 x 15.999 g/mol for O) + (3 x 1.008 g/mol for H) = 78.003 g/mol \( \text{mass of Al(OH)3} = \text{moles of Al(OH)3} \times \text{molar mass of Al(OH)3} \) \( \text{mass of Al(OH)3} = 0.1892 \mathrm{~mol} \times 78.003 \mathrm{~g/mol} \approx 14.76 \mathrm{~g} \) Therefore, approximately 14.76 grams of aluminum hydroxide are obtained when 14.2 grams of aluminum sulfide reacts with water.

Key concepts

Stoichiometry

Stoichiometry is like a recipe in chemistry. It helps us understand how different chemical reactants and products relate in a reaction. In our example with aluminum sulfide reacting with water, stoichiometry tells us the proportions needed. By using a balanced chemical equation, we know exactly how much of one substance will react with another or how much product will be formed.

When dealing with stoichiometry:

• Identify the balanced equation, which shows the relationship between reactants and products.
• Use the mole ratios to determine quantities. For instance, from the balanced equation, each mole of aluminum sulfide produces two moles of aluminum hydroxide.

Understanding stoichiometry allows you to predict the outcome of the reaction in terms of mass, enabling precise calculations.

Chemical Equation Balancing

Balancing a chemical equation ensures that the same number of each type of atom appears on both sides of the reaction. This is crucial because of the law of conservation of mass, which states that matter cannot be created or destroyed.

In our example, the unbalanced equation was:\[\text{Al}_2\text{S}_3 + \text{H}_2\text{O} \rightarrow \text{Al(OH)}_3 + \text{H}_2\text{S}\]To balance it, we adjusted coefficients:\[\text{Al}_2\text{S}_3 + 6\text{H}_2\text{O} \rightarrow 2\text{Al(OH)}_3 + 3\text{H}_2\text{S}\]This guarantees an equal number of aluminum, sulfur, hydrogen, and oxygen atoms on both sides.

Balancing equations involves trial and error, but understanding how atoms rearrange during reactions is key.

Molar Mass

Molar mass is the mass of one mole of a substance, measured in grams per mole (g/mol). It's essential for converting between grams and moles when solving chemistry problems.

To find molar mass, add up the atomic masses of all atoms in a compound. For instance, the molar mass of aluminum sulfide (Al₂S₃) is calculated by:

• Adding 2 times the atomic mass of aluminum (26.98 g/mol).
• Adding 3 times the atomic mass of sulfur (32.06 g/mol).

This gives us a molar mass of 150.16 g/mol for Al₂S₃.

Knowing this, you can convert given grams of a substance into moles to use in stoichiometric calculations, such as determining how many moles of a product you will get from a reaction.