# Problem: Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al2S3 + H2O ⟶Al(OH)3 + H2S [unbalanced]How many grams of the excess reactant remain?

###### FREE Expert Solution
• Using stoichiometry and a balanced equation, we can find the moles and mass of H2S using the molar masses and mass of reactants given

• The limiting reactant will tell how much moles of H2S will be produced

• Balancing the reaction, balance Al first (2 on both sides):

Al2S+ H2O ⟶ 2Al(OH)3 + H2S

• Balance S (3 on both sides)

Al2S+ H2O ⟶ 2Al(OH)3 + 3H2S

• Balance H and O (6 O and 10 H atoms on both sides)

Al2S+ 6H2O ⟶ 2Al(OH)3 + 3H2S (balanced equation)

• Calculating for the molar masses:

MM of H2S

H - 1.01 (2) = 2.02

S - 32.07 (1) = 32.07

2.02 + 32.07 = 34.09 g/mol

MM of Al2S3:

Al - 26.98 (2) = 53.96

S - 32.07 (3) = 96.21

53.96 + 96.21 = 150.17 g/mol

MM of H2O :

H - 1.01 (2) = 2.02

O - 16 (1) = 16

16 + 2.02 = 18.02 g/mol

• Calculating for the moles of H2S produced by the reactants:

95% (408 ratings) ###### Problem Details

Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water: Al2S3 + H2O ⟶Al(OH)3 + H2S [unbalanced]

How many grams of the excess reactant remain?