The complete combustion of ethanol, c2h5oh(l), to form h2(l)(g) and co2(g) at constant pressure releases 1235 kj of heat per mole c2h5oh a) write a balanced thermo chemical equation for this reaction b) if 5.25×10^24 molecules of co2 were produced during this reaction, how many kj was released? c) if 7525 kj were released during this reaction during this reaction, how many grams of the oxygen molecule(o2) were consumed?

Q1)
Ethanol is an alcohol that undergoes combustion in the presence of O₂ in air. Combustion yields water and carbon dioxide.
we are asked to write a balanced thermochemical equation, where the sum of the masses of products needs to be equal to the sum of the masses of the reactants.
when heat is released in the reaction, its known as an exothermic reaction and written with a negative sign. If heat is absorbed in the reaction, its called an endothermic reaction and written with a positive sign.
Combustion of ethanol is exothermic.
the balanced thermo chemical reaction for combustion of ethanol is as follows 
C₂H₅OH(l) + 3O₂(g) ---> 2CO₂(g) + 3H₂O(l)                 ΔH = -1235 kJ/mol

Q2)
Stoichiometry of C₂H₅OH to CO₂ is 1:2
When 1 mol of ethanol reacts, then 2 mol of CO₂ is produced , during this reaction 1235 kJ is released 
1 mol of CO₂ has 6.022 x 10²³ CO₂ molecules
Therefore 2 mol of CO₂ - 6.022 x 10²³ CO₂ molecules
When 6.022 x 10²³ CO₂ molecules are produced - 1235 kJ released 
When 5.25×10²⁴ CO₂ molecules are produced - 1235 x 5.25×10²⁴ / (6.022 x 10²³) = 10 766.8 kJ of heat is released

Q3)
Energy of 1235 kJ is released when 1 mol of ethanol reacts
For 7525 kJ of energy to be released - 1 x 7525 /1235 mol  = 6.09 mol of ethanol is required 
stoichiometry of C₂H₅OH to O₂ is 1:3
1 mol of C₂H₅OH reacts with 3 mol of O₂
amount of O₂ moles that react with 6.09 mol of C₂H₅OH - 6.09 x 3 = 18.28 mol
Mass of O₂ reacted - 18.28 mol x 32 g/mol = 585.0 g 

Yes you are correct use the scientific method

Ionic compounds dissociate into component ions dissolved into a solution creating substances called "electrolytes".hope my answer has you : )