Sucrose (C12H22O11, table sugar) is oxidized in the body by O2 via a complex set of reactions that ultimately produces CO2 (g) and H2O (g) and releases 5.16x10^3 kJ/mol sucrose.

a) Write a balanced thermochemical equation for this reaction. Make sure to include the physical state of each reactant and product........>>>>>>..........

b)Enter the answer for Delta Hrxn ........................answer in SCIENTIFIFIC NOTATION in kJ Moles
c) How much heat is released per gram of sucrose oxidized? ...................... answer in kJ/g

is b.....-5.16 x 10^3
is .c..............15.1 kJ/g ??? thank you.

C12H22O11(s) + 12O2(g) ==> 12CO2(g) + 11H2O(g) dH = -5.16E-3 kJ

b is -5.16E+3 kJ/mol
c is right

a) The balanced thermochemical equation for the oxidation of sucrose is:

C12H22O11 (s) + 12 O2 (g) -> 12 CO2 (g) + 11 H2O (g)

b) To find Delta Hrnx (the change in enthalpy of the reaction), we need to look at the given value of 5.16x10^3 kJ/mol. The stoichiometric coefficient of sucrose (C12H22O11) in the balanced equation is 1, meaning it represents 1 mole of sucrose. Therefore, the Delta Hrnx is indeed -5.16 x 10^3 kJ/mol.

c) To calculate the heat released per gram of sucrose oxidized, we need to convert the Delta Hrnx value into kJ/g. First, we need to determine the molar mass of sucrose.

C12H22O11 has a molar mass of:

12(12.01 g/mol) + 22(1.01 g/mol) + 11(16.00 g/mol) = 342.3 g/mol

Now, we can convert kJ/mol to kJ/g by dividing the Delta Hrnx value by the molar mass of sucrose:

(-5.16 x 10^3 kJ/mol) / (342.3 g/mol) = -15.1 kJ/g

Therefore, the correct answer for part c) is -15.1 kJ/g, not 15.1 kJ/g. Note the negative sign indicates that heat is released during the reaction.