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 )

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.