A chemist wants to make aluminum sulfide according to the following reaction:
3(NH4)2S (aq) + 2Al(NO3)3 (aq) → Al2S3 (s) + 6AH4NO3 (aq)
If he has an excess of ammonium sulfide and 315.0 mL of a 2.08 M solution of Al(NO3)3 how many grams of aluminum sulfide can he make?
- hcbiochemLv 71 month ago
Moles Al(NO3)3 = 0.3150 L X 2.08 mol/L = 0.655 mol Al(NO3)3
Mass Al2S3 = 0.655 mol Al(NO3)3 X (1 mol Al2S3 / 2 mol Al(NO3)3) X 150.16 g/mol = 49.2 g Al2S3