Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Chemistry Equation?

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? 

1 Answer

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  • 1 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

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