The heat released by one mole of sugar from a bomb calorimeter experiment is 5648 kJ/mol. The balanced chemical reaction equation is C12H22O11(s) + 12 O2(g) = 12 CO2(g) + 11 H2O(l) Calculate the enthalpy of combustion per mole of sugar. (answer in kJ/mol) A) 342.3 kJ/mol Eliminate B) 2824 kJ/mol C) 5648 kJ/mol D) 7250 kJ/mol

The answer to your question is the letter C) 5648 kJ/mol

Explanation:

Data

                C₁₂H₂₂O₁₁  +  12 O₂  ⇒   12 CO₂  +  11 H₂O

H° C₁₂H₂₂O₁₁ = -2221.8 kJ/mol

H° O₂ = 0 kJ / mol

H° CO₂ = -393.5 kJ/mol

H° H₂O = -285.8 kJ/mol

Formula

ΔH° = ∑H° products - ∑H° reactants

Substitution

ΔH° = 12(-393.5) + 11(-285.8) - (-2221.8) - (0)

ΔH° = -4722 - 3143.8 + 2221.8

Result

ΔH° = -5644 kJ/mol

the temperature of the gas, in k, not celsius.