Question

Calculate the no. of grams of Al2S3 which can be prepared by reaction of 20 grams of Al and 30 grams of sulphur.How much non limiting reactant is in access?

Answer 1

Answer:

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Explanation:

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.