# Chemistry

K+(aq) + Al3+(aq) + 2SO42-(aq) + 2H2O(l)→KAl (SO4)2 * 12H2O(s)

Al is limiting and 1.007 g
Find the theoretical yield and percent yield
I think the answer is 17 and 40, respect.
I just need help figuring it out

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1. Is that 1.007 g Al or 1.007 something different? Is that 12H2O on the left or 2 H2O? I assume 12.
mols Al = 1.007/atomic mass Al
Use the coefficients to change mols Al to mol KAl(SO4)2.12H2O
Now convert to grams. g = mols x molar mass. This is the theoretical yield (TY) and I obtained 17.7 g if your talking KAl(SO4)2.12H2O
%yield = (actual yield/TY)*100 = ? You don't list an actual yield in the problem.

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