posted by sss .
calculate the number of gram of Al2S3 which can be prepared by the reaction of 20 gram of Al and 30 gram of sulpher . how much the non limiting reactant is in excess?
The following will solve any "single" stoichiometry problem yu have. I work the limiting reagent problems two times, once with each reagent.
Use 30 g S and all the Al you need; solve for moles Al2S3. Then use 20 g Al and all the S needed and solve for moles Al2S3. Both answers can't be right; the correct answer in limiting reagent problems is ALWAYS the smaller one. Take that and convert to moles.
To find how much of the other reagent is in excess just do another stoiochiometry problem.
20g of Al = 20/26.98 mol = 0.74 mol
30g of S = 30/32.07 mol = 0.94 mol
so, only min(.74/2,.94/3) = 0.31333 mol of Al2S3 can be formed:
.31333 mol Al2S3 = .61667*26.98g Al + .94*32.07 g S = 16.64g Al + 30g S
That leaves 3.36g Al unreacted.
It gives 'Limiting Reactant' relationship.