2 Al + 3 S -> Al2S3 :This is the given reaction?
If 83.7 grams of aluminum, Al, and 195.2 grams of sulfur, S, are put into a container and allowed to react according to the above equation, which substance(s) and how many grams of each would be present in the container after the reaction is complete?
Atualizada:I tried to work it out and got 192.05?? I don't think I get it.
Answers & Comments
Verified answer
You just have to determine which is the limiting and excess reactant.
For Al:
83.7 grams Al x (1 mole Al / 26.98 grams Al) x (1 mole Al2S3 / 2 mole Al) = 1.55 mole Al2S3
For S:
195.2 grams S x (1 mole S / 32.06 grams S) x (1 mole Al2S3 / 3 mole S) = 2.03 mole Al2S3
Since Al is the limiting reactant, no Al would remain after the reaction is complete.
So, we just have to solve for the excess in S.
To solve for it, the equation we used in S [195.2 grams S x (1 mole S / 32.06 grams S) x (1 mole Al2S3 / 3 mole S) = 2.03 mole Al2S3] would be altered a bit.
Instead of 195.2 grams S, we make it Y, the variable that we are going to solve.
Instead of 2.03 mole Al2S3, we would change it to 1.55 mole Al2S3
Then just solve for Y.
Once you have solved for it, subtract Y from 195.2 grams S to get the excess of S.