Chem homework help urgent

Question

For the following reaction, 32.3 grams of aluminum oxide are allowed to react with 97.5 grams of sulfuric acid .

aluminum oxide(s) + sulfuric acid(aq)  aluminum sulfate(aq) + water(l)

What is the maximum amount of aluminum sulfate that can be formed?  grams

What is the FORMULA for the limiting reagent?

What amount of the excess reagent remains after the reaction is complete? 

Answer 1

So, this is very much like the other questions you have posted, and I have answered. I'm hoping that you are getting the "gist" of how to do these. Otherwise....well....

Al₂O₃ + 3H₂SO₄ ==> Al₂(SO₄)₃ + 3H₂O

32.3 g Al2O3 x 1 mol/102 g = 0.317 moles

97.5 g H2SO4 x 1mol/98 g = 0.995 moles

Al2O3 is LIMITING

Max. amount of Al2(SO4)3 = 0.317 mol Al2O3 x 1 mol Al2(SO4)3/1mol Al2O3 = 0.317 mol Al2(SO4)3

Since the question doesn't ask for mass, it just asks for maximum amount, this can be an answer. If you want the mass, multiply moles by molar mass. The question is poorly written.

I've done the rest of this question several times before, so you should be able to finish it up.