calculate the number of grams of AL2S3 which can be prepared by the re

1.	calculate the number of grams of AL2S3 which can be prepared by the reaction of 20g of (Al) and 30g of (sulphur).How much the non_limiting reactant is in excess?
Answer» Answer: 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 = .6166726.98g Al + .9432.07 G S = 16.64g Al + 30g S That leaves 3.36g Al unreacted

calculate the number of grams of AL2S3 which can be prepared by the reaction of 20g of (Al) and 30g of (sulphur).How much the non_limiting reactant is in excess?

Answer»

Answer:

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