I don't know why it was marked wrong. This is what it said: Where is the 5.64 coming from? (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.
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
To understand where the value of 5.64 x 10^3 kJ comes from, you need to refer to the given information. The information states that the reaction of oxidizing 1 mole of sucrose (C12H22O11) releases 5.16 x 10^3 kJ of energy.
To find out how many moles of sucrose are equivalent to 5.64 x 10^3 kJ, you can set up a conversion factor using the given information:
1 mol C12H22O11 releases 5.16 x 10^3 kJ
Now, you can use this conversion factor to calculate how much heat is released per gram of sucrose oxidized. To do this, you need to convert grams of sucrose to moles of sucrose using its molar mass (342.3 g/mol). Then, you can use the conversion factor to find the amount of heat released.
Let's calculate it step by step:
Step 1: Convert grams of C12H22O11 to moles of C12H22O11
Assuming you have a certain mass of sucrose (let's say x grams), you can use the molar mass of sucrose (342.3 g/mol) to convert grams to moles:
moles of C12H22O11 = x g * (1 mol / 342.3 g)
Step 2: Use the conversion factor to find the amount of heat released
Now that you have moles of sucrose, you can use the conversion factor (1 mol C12H22O11 releases 5.16 x 10^3 kJ) to find the amount of heat released:
heat released = moles of C12H22O11 * (5.16 x 10^3 kJ / 1 mol)
Make sure to cancel out units properly and calculate the answer correctly. From the information given, it seems like b should be -5.16 x 10^3 kJ (negative because energy is being released) and c should be calculated based on the steps mentioned above to get 15.1 kJ/g.