Suppose 316.0 g aluminium sulphide reacts with 493.0 g of water. How much of the excess reactant is left over at the end of the reaction? Al2S3+H2O→Al(OH)3+H2S
The correct option is A 265.4 g The unbalanced equation is: Al2S3+H2O→Al(OH)3+H2S The balanced equation is: Al2S3+6H2O→2Al(OH)3+3H2S Number of moles of Al2S3=316.0g150g/mol=2.107mol Number of moles of H2O=493.0g18g/mol=27.388mol
For Al2S3 Initial number of molesstoichiometric coefficient=2.1071=2.1044
For H2OInitial number of molesstoichiometric coefficient=27.3886=4.564
Al2S3 is the limiting reagent.
To find the number of moles of H2O required to consume 2.107 mol of Al2S3: