J.R. S. answered 11/11/22
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0.65 mols Ba / 3 = 0.2
0.50 mols Al2(SO4)3 / 1 = 0.5
Excess reactant is Al2(SO4)3
Massiah M.
asked 11/10/223 Ba + Al2(SO4)3 → 2 Al + 3BaSO4
If you have .65 moles of Ba and .50 mole of Al2(SO4)3, what is the excess reactant?
J.R. S. answered 11/11/22
Ph.D. University Professor with 10+ years Tutoring Experience
0.65 mols Ba / 3 = 0.2
0.50 mols Al2(SO4)3 / 1 = 0.5
Excess reactant is Al2(SO4)3
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