Question

Calculate the maximum numbers of moles and grams of H₂S that can form when 147.3 g of aluminum sulfide reacts with 143.0 g of water:

Al₂S₃ + H₂O → Al(OH)₃ + H₂S [unbalanced]

What mass of the excess reactant remains?

Answer 1

The balanced chemical reaction between Al2S3 and H2O is shown below,

Al2S3 + 6H2O = 2Al(OH)3 + 3H2S

Let is first calculate the number of moles of Al2S3 and moles of H2O used in the reaction,

The molar mass of Al2S3 and H2O are 150.16 g/mol and 18 g/mol, respectively

The number of moles of Al2S3 = Weight of Al2S3/ molar mass of Al2S3

                   = 147.3 g / 150.16 g/mol = 0.981 moles

The number of moles of H2O   = Weight of H2O / molar mass of H2O

       = 143 g / 18 g/mol = 7.94 mol

From the balanced chemical equation it is clear that 1 mole of Al2S3 requires 6 moles of H2O.

Therefore, the ideal ratio of number of moles of Al2S3 and H2O = 1/6 = 0.166

The actual ratio of number of moles of Al2S3 and H2O = 0.981 / 0.794 = 0.123

As the actual ratio of number of moles of Al2S3 and H2O is less than the ideal ratio, H2O is used in excess and Al2S3 is the used less in the reaction which is a limiting reagent. The mass of H2S formed will depend on the number of moles of Al2S3 used.

From the balanced chemical equation it is clear that 1 moles of Al2S3 produces 3 mole of H2S. Therefore, we can consider,

1 moles of Al2S3   = 3 mole of H2S

For, 0.981 moles of Al2S3 = 0.981 x 3 = 2.943 moles of H2S

The mass of H2S formed = Number of moles of H2S formed x molecular weight of H2S in g/mol

The molecular weight of H2S is 34.1 g/mol

The mass of H2S formed = 2.943 mol x 34.1 g/mol

= 100.35 g

Therefore 100.35 g of H2S will be formed.

As Al2S3 is getting consumed completely in the reaction, the excess reagent is H2O.

The excess mass of H2O = Mass of H2O used in g - Mass of H2O consumed in g

The mass of H2O consumed in g can be calculated from number of moles of Al2S3 used. As, from the balanced chemical equation 1 moles of Al2S3 consumes 6 mole of H2O.

Therefore, we can consider,

1 moles of Al2S3 used = 6 mole of H2O consumed

For, 0.981 moles of Al2S3 = 0.981 x 6 = 5.88 moles of H2O consumed

Mass of H2O consumed = number of moles of H2O consumed x molecular weight of H2O

in g/mol

= 5.88 mol x 18 g/mol

= 105.94 g

The excess mass of H2O = 143 g – 105.94 g = 37.06 g

Therefore, 37.06 g of excess reactant will remain after completion of reaction