Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide by the following reaction
Al₂S₃ + 6 H₂O $\rightarrow$ 2Al(OH) ₃ + 3H₂S
If 25 g of aluminum sulfide is reacted with 25 g of water how many moles of hydrogen sulfide will be formed?

Question

Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide by the following reaction
Al₂S₃ + 6 H₂O

\rightarrow

2Al(OH) ₃ + 3H₂S
If 25 g of aluminum sulfide is reacted with 25 g of water how many moles of hydrogen sulfide will be formed?

Quizplus · Accepted Answer

First, calculate the moles of aluminum sulfide (Al2S3) and water (H2O) used. The molar mass of Al2S3 is approximately 150.16 g/mol, and for H2O, it is 18.015 g/mol. For 25 g of Al2S3, moles = 25 g / 150.16 g/mol ≈ 0.166 moles. For 25 g of H2O, moles = 25 g / 18.015 g/mol ≈ 1.387 moles. The reaction shows that 1 mole of Al2S3 reacts with 6 moles of H2O to produce 3 moles of H2S. Since Al2S3 is the limiting reagent, it will completely react, and the amount of H2S produced will be directly proportional to the amount of Al2S3. Therefore, 0.166 moles of Al2S3 will produce 0.166 * 3 = 0.498 moles of H2S, which is approximately 0.50 moles.

Aluminum Sulfide Reacts with Water to Form Aluminum Hydroxide and Hydrogen