2Al ==> Al2(SO4)3.K2SO4.24H2O
The above isn't fully balanced but I've balanced the only part that matters.
mols Al = grams/molar mass = approx 0.0074
Using the coefficients in the equation convert mols Al to mols of alum. That's 0.0074 x (1 mol alum/2 mol Al) = 0.0074 x 1/2 = 0.0037
g alum produced = mols alum x molar mass alum.
NOTE: Some write the formula for alum as the above; others write it as half that (that is KAl(SO4)2.12H2O but that will not change the answer you get.)
Calculate the theoretical yield of alum if 0.200g of Aluminum was reacted with excess KOH and H2SO4?
I don't even know how to start it. Please help. Also I wanted to say thank you for the help I've gotten from this website it has helped me a lot. I feel like I'm understanding the material better ever since I found it.
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