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How to calculate % of water and mass of water in my alum product?
Just did a lab where I synthesized alum from aluminum and I am stuck on these questions.
The overall reaction equation is this :
2Al + 2KOH + 4h2SO4 + 22(H2O) ------------> 2KAl(SO4)2 * 12H2O + 3H2
This means there are 12 moles of H2O in the product for every mole of aluminum right?
1.102g of aluminum gave me 17.82g of alum
So can anyone give me a tip on how to calculate the % of water and mass of water in my alum product? Been at it for quite a bit now and have no idea how to do these questions.
1 Answer
- JOHLv 68 years agoFavorite Answer
Determine the moles of your alum sample (molar mass = 474.4 g-mol)
n(KAl(SO4)2•12H2O) = 17.82 g / 474.4 g-mol = 0.03756 mol
Now determine the moles of water; there are 12 moles of H2O per mole of alum
n(H2O) = (12)(0.03756) = 0.4508 mol
Determine mass of H2O by multiplying by molar mass of H2O (18.02 g-mol)
(0.4508 mol)(18.02 g-mol) = 8.123 g
% of water by mass in your alum sample:
(8.123 / 17.82)(100) ≈ 45.58 %
