Question

Calculate the max # of moles and grams of H2S that can form when 143 g aluminum sulfide reacts with 128.8 g water?

a) ____mol H2S b) _____g H2S c) ____g excess reactant

Answer 1

Answer – We are given, mass of aluminium sulfide, Al₂S₃ = 143 g

Mass of water = 128.8 g

Reaction –

Al₂S₃ + 6 H₂O ------> 3 H₂S + 2 Al(OH)₃

Now convert the mass to moles of each given reactant-

Moles of Al₂S₃ = 143 / 150.16 g.mol^-1

                         = 0.952 moles

Moles of H₂O = 128.8 g / 18.015 g.mol^-1

                       = 7.15 moles

Calculate the limiting reactant –Calculate the moles of H₂S from both reactants

Moles of H₂S from moles of Al₂S₃

1 moles of Al₂S₃ = 3 moles of H₂S

So, 0.952 moles of Al₂S₃ = ?

= 2.86 moles of H₂S

Moles of H₂S from moles of H₂O

6 moles of H₂O = 3 moles of H₂S

So, 7.15 moles of H₂O = ?

= 3.57 moles of H₂S

So moles of H₂S lowest form the Al₂S₃, so Al₂S₃ is limiting reactant and H₂O is excess reactant

So, maximum number of moles of H₂S formed = 2.86 moles

Mass of H₂S = 2.86 moles * 34.081 g/mol

                     = 97.4 g of H₂S

Moles of H₂O reacted –

1 moles of Al₂S₃ = 6 moles of H₂O

So, 0.952 moles of Al₂S₃ = ?

= 5.71 mole of H₂O

So excess moles of H₂O = 7.15 – 5.71

                                      = 1.44 moles

Excess mass of reactant = 1.44 moles * 18.015 g/mol

                                       = 25.87 g