Theoretical yield of alum from ammonium sulfate and aluminum sulfate

[curlyfries4]

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.

 

[chemisttree]

Balance this first...

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

 

[Borek]

Note: you will need water on the LHS.

 

[curlyfries4]

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

 

[Borek]

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