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 is: Al2S3 (s) + 6H2O (l) -> 2Al(OH)3 (aq) + 3H2S (g) (b) Approximately 14.77 g of aluminum hydroxide are obtained from 14.2 g of aluminum sulfide.

Step-by-step solution

Write the balanced chemical equation

The unbalanced chemical equation for the reaction can be written as: Al2S3 (s) + H2O (l) -> Al(OH)3 (aq) + H2S (g) To balance the equation, we need to make sure there are the same number of each element on both the reactant side and the product side. After balancing, the equation becomes: Al2S3 (s) + 6H2O (l) -> 2Al(OH)3 (aq) + 3H2S (g)

Calculate the moles of aluminum sulfide

To calculate the moles of aluminum sulfide, we can use the given mass (14.2 g) and molar mass of aluminum sulfide, which is approximately 150 g/mol (26.98 g/mol for aluminum and 32.07 g/mol for sulfur). Moles of Al2S3 = (14.2 g) / (150 g/mol) = 0.0947 mol

Use the mole ratio to find the moles of aluminum hydroxide produced

From the balanced equation, we can see that 1 mole of Al2S3 produces 2 moles of Al(OH)3. So, with the available moles of aluminum sulfide: Moles of Al(OH)3 = Moles of Al2S3 × (2 moles of Al(OH)3 / 1 mole of Al2S3) = 0.0947 mol × 2 = 0.1894 mol

Calculate the mass of aluminum hydroxide produced

To find the mass of aluminum hydroxide produced, we can use the moles calculated in the previous step and the molar mass of aluminum hydroxide. The molar mass of Al(OH)3 is approximately 78 g/mol (26.98 g/mol for aluminum and 3 × 16 g/mol for oxygen plus 3 × 1 g/mol for hydrogen): Mass of Al(OH)3 = Moles of Al(OH)3 × Molar mass of Al(OH)3 Mass of Al(OH)3 = (0.1894 mol) × (78 g/mol) = 14.77 g So, from 14.2 g of aluminum sulfide, approximately 14.77 g of aluminum hydroxide are produced.