What mass of Al is produced when 0.500 mole of Al2S3 is
completely reduced with excess H2?
Answers
Answered by
5
Answer:
Not sure
The ratio between Al and Al2S3 is 2 : 1
Moles Al2S3 = 3.00 x 1 / 2 = 1.50
Mass Al2S3 = 1.50 mol x 150.155 g/mol = 225.2 g
adityapatil12102003:
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Answered by
1
Answer:
The mass of Al produced is 27 g.
Explanation:
we could consider the chemical reaction, (Al₂S₃⇔ 2 Al + 3S) or we could simply realize that since the Al would be written as a monatomic atom( a combination of the words "mono" and "atomic", and means "single atom") in the balanced equation and since there are 2 Al’s in the compound, there must be a 2:1 ratio between the Al₂S₃ and Al :
0.5 mole Al2S3 = 2 Al/ 1 Al2S3= 1 mole Al (27g/ 1 mol) = 27 g Al.
Hence, 27 g of Al is produced.
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