Sucrose, C12H22O11, is table sugar, and it has a ∆Hc of –5639.7 kJ/mol. Determine the ∆Hf of sucrose given

Question

∆Hf (H2O(l)) = –285.8 kJ/mol, and ∆Hf (CO2(g)) = –393.5 kJ/mol .
C12H22O11(s) + O2(g) → CO2(g) + H2O(l)

Answer 1

I don't know your definition of delta Hc but I don't know that it is needed to solve for dHf sucrose.
C12H22O11(s) + (35/2)O2(g) →12CO2(g) + 11H2O(l)
dHrxn = (n*dHf products) - (n*dHf reactants)
Plug in the values and solve. Post your work if you get stuck.