For the following reaction, 3.04 grams of sulfuric acid are mixed with excess aluminum oxide. The reaction yields 2.53 grams of aluminum sulfate. aluminum oxide (s) sulfuric acid (aq) aluminum sulfate (aq) water (l) What is the theoretical yield of aluminum sulfate

Theoretical yield of aluminum sulfate is 3.52 g

Explanation:   The balanced reaction of sulfuric acid with aluminum oxide is given by :

     →  

So  1 mole of aluminum oxide reacts with 3 moles of sulfuric acid to form 1 mole of aluminum sulfate  and 3 moles of water .  Here aluminum oxide is in excess   so product will be formed according to sulfuric acid .

Moles of sulfuric acid  =  mass / molar mass    =   3.04 g / 98.07 g/mole  

                                                                              =  0.0309 moles

As  we now that 3 moles of sulfuric  forms 1 mole of aluminum sulfate .

So  0.0309 moles of sulfuric acid forms   =   1 / 3 x  0.0309  

                                                                       =  0.0103 moles .

    Now  0.0103 moles of aluminum sulfate  =  0.0103 moles x 342.15 g/mole  

                                                                           =  3.52 g

      ( 342. 15 g /mole   =  Molar mass of aluminum sulfate )

the answer is the first option

explanation:

just googled it m8. not that hard

lol

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"