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Calculating theoretical yield of resulting alum from water dissolving help asap!

  1. Nov 9, 2008 #1
    calculating theoretical yield of resulting alum from water dissolving....help asap!

    a mass of 13.02g (NH4)2SO4 is dissolved in water. After the solution is heated, 27.22g of Al2(SO4)3*18H2O is added. calculate the theoretical yield of the resulting alum (formula is NH4+(superscript)Al3+(superscript)(SO4)2*12H2O) Hint:this is a limiting reactant problem.

    so far ive got 2Al+(NH4)2SO4+H2O yields Al2(SO4)3*18H2O + the formula of alum but im rly not sure if this is right...please hellppp meeee asap!thanks.
     
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  3. Nov 11, 2008 #2

    chemisttree

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    Re: calculating theoretical yield of resulting alum from water dissolving....help asa

    Balance this first....

    Al2(SO4)3*18H2O + (NH4)2SO4 -----> NH4Al(SO4)2·12H2O
     
  4. Nov 11, 2008 #3

    Borek

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    Re: calculating theoretical yield of resulting alum from water dissolving....help asa

    Note: you will need water on the LHS.
     
  5. Nov 11, 2008 #4
    Re: calculating theoretical yield of resulting alum from water dissolving....help asa

    but i dont know how to balance with hydrates that well...the limiting reactant side?
     
  6. Nov 11, 2008 #5

    Borek

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    Re: calculating theoretical yield of resulting alum from water dissolving....help asa

    Write hydrate as Al2(SO4)3(H2O)18. LH stands for left hand.
     
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