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Based on the balanced equation

Al2S3 + 6H2O → 2Al(OH)3 + 3H2S
calculate the number of excess reagent units remaining when 4 Al2S3 formula units and 30 H2O molecules react?

Molar Mass (g/mol)
Al2S3 150.16
H2O 18.015
Al(OH)3 75.361
H2S 34.082
Avogadro's No.
6.022×1023 mol-1

  • Chemistry - ,

    This is a limiting reagent problem. Step 1 is to determine the limiting reagent; i.e., is it Al2S3 or H2O? You determine this from looking at the coefficients in the balanced equation. Just pick one. I'll pick H2O first, then do Al2S3.
    The coefficients tell us that 6 units of H2O require 1 unit of Al2S3; therefore, 30 units of water (given in the problem) will require 7.5 units of Al2S3 and the way you get that is
    30 units H2O x (1 unit Al2S3/4 units H2O) = 7.5 units Al2S3. Note that I arranged the factor (1 unit Al2S3/4 units H2O) so that the factor converts units of H2O to units of Al2S3 by making the units H2O in the first term cancel with units H2O in the denominator of the factor). So, do we have 7.5 units of Al2S3. No, the problem says we have 4 so we know Al2S3 is the limiting reagent BUT let's do it the same way and check it.
    4 units Al2S3 x (6 units H2O/1 unit Al2S3) = 24 units H2O. Do we have 24? Yes, the problem says we have 30 so Al2S3 indeed is the limiting reagent, it will require 24 units of H2O to use all 4 units of Al2S3, which means 30-24=6 units of water will remain after the reaction is complete. Check my work.

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