What mass of Al is produced when 0.500 mole of Al2S3 is
completely reduced with excess H2?
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What mass of Al is produced when 0.500 mole of Al2S3 is
completely reduced with excess H2?
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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