I don't know why it was marked wrong. This is what it said: Where is the 5.64 coming from? Is this a typo?(thank you)
HINT
To balance the equation, make sure that the number of each atom on the reactant side is the same as the number on the product side. Should the enthalpy change for this reaction have a positive sign or a negative sign? From the balanced thermochemical equations, how many moles of C12H22O11 (sucrose) are thermochemically equivalent to 5.64 x 10^3 kJ? Use this equivalent quantity as a conversion factor to calculate how much heat is released per gram of sucrose oxidized . Make sure all your units cancel out and the signs are correct in your calculations.
THIS WAS THE QUESTION:
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.•chemistry - DrBob222, Sunday, October 18, 2015 at 1:32pm
C12H22O11(s) + 12O2(g) ==> 12CO2(g) + 11H2O(g) dH = -5.16E-3 kJ
b is -5.16E+3 kJ/mol
c is right
I would help but I don't know how...I'm sorry. Maybe Dr. Bob can help you when he wakes up :)
Actually, it was a typo...dr bob was right. Thank you, dr bob!
The 5.64 x 10^3 kJ mentioned in the question seems to be a typo. The correct value is 5.16 x 10^3 kJ/mol, as mentioned in the balanced equation.
To answer the question step-by-step:
a) The balanced thermochemical equation for the reaction is:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(g)
b) The enthalpy change for this reaction, ΔHrxn, is -5.16 x 10^3 kJ/mol. The negative sign indicates that the reaction is exothermic, meaning it releases heat.
c) To calculate how much heat is released per gram of sucrose oxidized, we need to use the given thermochemical information. The 5.16 x 10^3 kJ/mol represents the heat released when 1 mole of sucrose is oxidized.
To convert this quantity to kJ/g, we need to convert the grams of sucrose to moles first. The molar mass of sucrose (C12H22O11) is approximately 342.3 g/mol.
Now we can set up the conversion:
(5.16 x 10^3 kJ/mol) / (342.3 g/mol) = 15.08 kJ/g (rounded to two decimal places)
Therefore, the answer for part c) is approximately 15.08 kJ/g.
The 5.64 mentioned in the question is not a typo, but rather a conversion factor that will be used to calculate the heat released per gram of sucrose oxidized.
To answer the question, we first need to write the balanced thermochemical equation for the reaction. The given information tells us that sucrose (C12H22O11) is oxidized by O2 in a complex set of reactions, resulting in the production of CO2 and H2O, and releasing 5.16x10^3 kJ/mol of sucrose.
The balanced thermochemical equation for the reaction is:
C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(g)
Now, let's move on to answering the specific parts of the question:
a) As mentioned above, the balanced thermochemical equation is C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(g).
b) The enthalpy change for this reaction, denoted as ΔHrxn, can be found by looking at the given information. It states that 5.16x10^3 kJ of heat is released per mole of sucrose. Therefore, the answer is -5.16x10^3 kJ/mol (a negative sign indicates the release of heat).
c) To calculate how much heat is released per gram of sucrose oxidized, we need to convert the given information from moles to grams. The question provides the conversion factor of 5.64x10^3 kJ/mol sucrose.
To use this conversion factor, we need to determine the molar mass of sucrose. The molar mass of C12H22O11 can be calculated using the atomic masses of carbon, hydrogen, and oxygen.
Now, we can use the conversion factor and molar mass to find the heat released per gram of sucrose oxidized:
(5.64x10^3 kJ/mol) / (molar mass of sucrose in grams) = heat released per gram of sucrose oxidized
Without knowing the molar mass of sucrose, it is not possible to calculate the final answer of 15.1 kJ/g.
I hope this explanation helps. Please let me know if you have any further questions.