Calculating theoretical yield of resulting alum from water dissolving help asap

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To calculate the theoretical yield of alum from the given reaction, first balance the equation, including the water molecules from the hydrate. The reaction involves 13.02g of (NH4)2SO4 and 27.22g of Al2(SO4)3*18H2O, with water also needed on the reactant side. Identifying the limiting reactant is crucial for accurate yield calculations. Once the equation is balanced, use stoichiometry to determine the theoretical yield of NH4Al(SO4)2·12H2O. Properly accounting for the hydrates and balancing the reaction will lead to the correct yield calculation.
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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 I've got 2Al+(NH4)2SO4+H2O yields Al2(SO4)3*18H2O + the formula of alum but I am rly not sure if this is right...please hellppp meeee asap!thanks.
 
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Balance this first...

Al2(SO4)3*18H2O + (NH4)2SO4 -----> NH4Al(SO4)2·12H2O
 


Note: you will need water on the LHS.
 


but i don't know how to balance with hydrates that well...the limiting reactant side?
 


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

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