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

The balanced chemical equation for the reaction between aluminum sulfide (Al2S3) and water (H2O) to form aluminum hydroxide (Al(OH)3) and hydrogen sulfide (H2S) is: \(Al2S3 + 6H2O \rightarrow 2Al(OH)3 + 3H2S\). From 14.2 g of aluminum sulfide, 14.8 g of aluminum hydroxide are obtained.

Step-by-step solution

1. Write the balanced chemical equation

The given reaction is between aluminum sulfide (Al2S3) and water (H2O), which forms aluminum hydroxide (Al(OH)3) and hydrogen sulfide (H2S). The chemical equation for this reaction is: \[Al2S3 + 6H2O \rightarrow 2Al(OH)3 + 3H2S\] This equation is balanced as there are equal numbers of each element on both sides.

2. Calculate the moles of aluminum sulfide

To determine the mass of aluminum hydroxide formed, we need to know the amount of aluminum sulfide in moles. We are given 14.2 g of aluminum sulfide. The molar masses of the elements are: Aluminum (Al) = 27 g/mol Sulfur (S) = 32 g/mol The molar mass of aluminum sulfide (Al2S3) is equal to the sum of the molar masses of its constituent elements: \(1 \times 2\times 27 ~g/mol + 1 \times 3\times 32 ~g/mol = 150 ~g/mol\) Now, we can find the moles of aluminum sulfide: \(\frac{14.2 ~g}{150 ~g/mol} = 0.0947 ~mol\)

3. Use stoichiometry to find moles of aluminum hydroxide

The stoichiometry of the balanced chemical equation tells us the relationship between the moles of reactants and products. From the equation, 1 mole of aluminum sulfide forms 2 moles of aluminum hydroxide: \(Al2S3 + 6H2O \rightarrow 2Al(OH)3 + 3H2S\) So, for every mole of aluminum sulfide, twice that number of moles of aluminum hydroxide is formed. To find the moles of aluminum hydroxide formed from the given amount of aluminum sulfide, multiply the moles of aluminum sulfide by 2: \(0.0947~ mol \times 2 = 0.1894~ mol\)

4. Calculate the mass of aluminum hydroxide

We have found the moles of aluminum hydroxide formed in the reaction. Now, we can calculate the mass of the aluminum hydroxide. The molar mass of aluminum hydroxide (Al(OH)3) is equal to the sum of the molar masses of its constituent elements: \(1 \times 27 ~g/mol + 3 \times 16 ~g/mol + 3 \times 1 ~g/mol = 78 ~g/mol\) Now, we can find the mass of aluminum hydroxide: \(0.1894 ~mol \times 78 ~g/mol = 14.8 ~g\)

5. Answer to part (b)

14.8 grams of aluminum hydroxide are obtained from 14.2 grams of aluminum sulfide in the given reaction.