Given the balanced equation representing a reaction:
4al(s) + o2(g) --> 2al2o3(s)
how many moles of al(s) react completely with 4.50 moles of o2(g)to produce 3.00 moles of al2o3(s)?
(1) 1.50 mol (2) 2.00 mol
(3) 6.00 mol (4) 4.00 mol

4Al(s) + 3O2(g) --> 2Al2O3(s)    This is the balanced.
From the equation:

 4 moles of Al required 3 moles of O2 to produce 2 moles of Al2O3
 
3 moles of O2 reacted with 4 moles of Al to produce 2 moles of Al2O3
1 mole of O2 reacted with 4/3 moles of Al to produce 2/3 moles of Al2O3  (Divide by 3)
4.5 moles of O2 reacted with (4/3 *4.5) moles of Al to produce (2/3*4.5) moles of Al2O3

4.5 moles of O2 reacted with 6moles of Al to produce  3moles of Al2O3

(3) is the answer.  6 mol of Al.

Sorry i dont understand the question can you rephase it?