Moles of Potassium Alum in 2KAl(SO_4)_2 * 12H_2O

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The discussion clarifies the molar composition of Potassium Alum, specifically the formula 2KAl(SO4)2 * 12H2O. It establishes that this formula represents 1 mole of Potassium Alum, despite containing 2 moles of Potassium and Aluminum. The confusion arises from the interpretation of the formula, where the presence of the coefficient '2' applies to the entire structure, indicating a 1:1 ratio between moles of Aluminum and Potassium Alum. Thus, 2 moles of Aluminum react with 2 moles of Potassium Alum.

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Hi guys,

I know the formula for Potassium Alum is:

2KAl(SO_{4})_{2} * 12H_{2}O

Now does this mean that there are 2 Moles of Potassium Alum or 24 moles of Potassium Alum if you include the 12 water molecules? I know when you calculate the molar mass of Potassium Alum, you need to add in the molar mass of the 12 water molecules.

Thanks
 
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It means there are 2 moles of potassium alum in that formula as written. The '*' just means that each molecule set 2KAl(SO_{4})_{2} is surrounded by 12 molecules of water. But if that is the exact formula for potassium alum then 2KAl(SO_{4})_{2} * 12H_{2}O is only 1 mole. There are 2 moles of potassium and 2 moles of aluminum, but that entire structure is only 1 mole of potassium alum.
 
Hi dmoravec, thanks for the reply. I am just confused on how there can be only 1 mole for that whole structure.

I just need help on converted moles of aluminum to moles of alum.

2K^+ + 2Al^+3 + 4SO_{4}^-2 + 12H_{2}O --> 2KAl(SO_{4})_{2} * H_{2}O

Now does this mean that 2 moles of Aluminum react with 2 moles of Potassium Alum or just 1 mole, as you mentioned above, of Postassium Alum?

Thanks again
 
well, its a bit nitpicky and just depends on naming but it could be either depending on how you define 1 molecule of potassium alum and how exactly the right side of the equation is written. If potassium alum is defined as 2KAl(SO_{4})_{2} * 12H_{2}O then it would be 2 moles of aluminum equating with 1 mole of potassium alum. If potassium alum is defined as KAl(SO_{4})_2*12H_{2}O then it would be 2 moles of aluminum equating with 2 moles of potassium alum.

(the difference lies in the formula above being (2KAl(SO_{4})_2*12H_{2}O) or
2(KAl(SO_{4})_2*12H_{2}O))
 
Thanks dmoravec...the formula is actually written as the latter, with the 2 on the outside of the entire formula, so it must be 2 moles of aluminum reacting with 2 moles of potassium alum, in other words a 1:1 ratio. Thanks again, it was a big help
 

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