How many grams of Al2S3 can be formed from the reaction of 108.00 grams of Al with 5.00 grams of S?

Answer 1

The answer i got is 12.33 grams of Al2S3. Below i will try to show the steps i used:

n=moles

m=mass (grams)

M= molecular weight (from Periodic Table)

2Al + 3S --> Al2S3

nAl= m/M = 9/13 = 0.692307 moles of Al

nS= m/M = 8/16 = 0.5 moles of S

Limiting Reagent: 1.5 moles of S required for every mole of Al (3:2 ratio)

This means that there should be about 1.04 moles of S to completely use up all the Al. Since there is less than this amount of S present (only 0.5 moles), S is the limiting reagent and should be used in the mole calculation of the product

Amount of product: 0.5 moles S x (1 mol Al2S3/ 3 mol S) = 0.16666 moles of Al2S3

nAl2S3 = m/M

Molecular weight of Al2S3 = (2 x 13) + (3 x 16) = 74

0.16666 moles Al2S3 = m/74

m = 74 x 0.16666 = 12.33 grams of Al2S3

Therefore, 12.33 grams of Al2S3 is formed in this reaction

( hopefully this is right :P )

Answer 2

7,8 g of aluminium sulfide are obtained.