Table sugar consists primarily of sucrose, C12H22O11(s). The standard enthalpy of combustion for sucrose is the ΔHo for the reaction:

C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)
Calculate this ΔHo. Give your answer in units of kJ to five significant figures. Use the following values of ΔHof (in kJ/mol): CO2(g) = –393.51; H2O(l) = –285.83; C12H22O11(s) = –2221.2.

To calculate the standard enthalpy of combustion (ΔHo) for the reaction, we need to use the given values of ΔHof (standard enthalpy of formation) for CO2(g), H2O(l), and C12H22O11(s).

Given values:
ΔHof for CO2(g) = -393.51 kJ/mol
ΔHof for H2O(l) = -285.83 kJ/mol
ΔHof for C12H22O11(s) = -2221.2 kJ/mol

The reaction equation for the combustion of sucrose is:
C12H22O11(s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l)

The ΔHo for the reaction is the sum of the standard enthalpies of formation of the products minus the sum of the standard enthalpies of formation of the reactants. Since there are 12 CO2 and 11 H2O in the products, the enthalpy change can be calculated as:

ΔHo = (12 × ΔHof of CO2) + (11 × ΔHof of H2O) - (1 × ΔHof of C12H22O11)

Substituting the given values:
ΔHo = (12 × -393.51) + (11 × -285.83) - (-2221.2)

Calculating the equation:
ΔHo = -4722.12 - 3144.13 + 2221.2

ΔHo = -5645.05 kJ

Therefore, the ΔHo for the combustion of sucrose is approximately -5645.05 kJ.