Question

(6 points) Calculate the maximum number of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Show work for full poi

Answer 1

Answer:107.1 g, 124.1 g

Explanation:

The equation of the reaction is;

Al2S3(s) + 6H20(l) ----> 2Al(OH)3(s) + 3H2S(g)

Hence;

For Al2S3

Number of moles= reacting mass/molar mass

Number of moles = 158g/150gmol-1 =1.05 moles

If 1 mole of Al2S3 yields 3 moles of H2S

1.05 moles of Al2S will yield

1.05 × 3/1 = 3.15 moles

Mass of H2S = 3.15moles × 34 gmol-1 = 107.1 g

For water

Number of moles of water = 131g/18gmol-1= 7.3 moles

6 moles of water yields 3 moles of H2S

7.3 moles of water will yield 7.3 × 3/6 = 3.65 moles of H2S

3.65 moles × 34 gmol-1 =124.1 g