Question

Worksheet for quiz 2 (chapters 2 and 3) 1. (3 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: ALS; + H2O + Al(OH)3 + H2S (unbalanced] How many grams of the excess reactant remain?

Answer 1

after balancing the equation,

Al₂S_​​​3 + 6H₂O $\rightarrow$ 2 Al(OH)₃ + 3 H_{​​​​​​2}​​​​​S

So, 1 mole of Al₂​​​​S_​​​3 ​​​​​​will react with 6 moles of H₂O

i.e., 150.16g of Al₂S_​​​3​​ will react with 108g(18×6) of H₂O

So, 158g of Aluminium sulfide will react with=(108÷150.16)×158 = 113.6g of H₂O

Here the limiting reagent is Aluminium sulfide.

So the excess reagent i.e., H₂O = Given mass of excess reagent - required mass of that reagent = 131 - 113.6 = 17.4g of H₂O

1mole ie, 150.16g of aluminium sulfide will give 3 moles of H₂S

158 ie, (158/150.16 = 1.05 moles) of Al₂S₃ will give= 3×1.05= 3.15 moles of H₂S

1 mole os H₂S = 34.1g

3.15 molesof H₂S = no of moles × molar mass = 3.15 × 34.1 =107.41g of H₂S