The value of ΔH° for the reaction below is - 6535 kJ. __________ kJ of heat are released in the combustion of 16.0 g of

C6H6(l)?
2C6H6(l) + 15O2(g) → 12CO2(g) + 6H2O (l)

6535 kJ/2 mol C6H6 = ? kJ/1 mol C2H6.

mols C2H6 in 16.0g = grams/molar mass
Then
kJ heat released = kJ/mol x # mols in 16.0g = ?

To determine the amount of heat released in the combustion of 16.0 g of C6H6(l), we can use the given value of ΔH° for the reaction and apply a simple stoichiometric calculation.

First, we need to find the molar mass of C6H6 (benzene). The molar mass of carbon (C) is approximately 12.01 g/mol, and the molar mass of hydrogen (H) is approximately 1.01 g/mol. Since there are 6 carbon atoms and 6 hydrogen atoms in one molecule of C6H6, the molar mass of C6H6 is:

(6 * 12.01 g/mol) + (6 * 1.01 g/mol) = 78.11 g/mol

Next, we can calculate the moles of C6H6 by dividing the mass by the molar mass:

moles of C6H6 = mass of C6H6 / molar mass of C6H6
moles of C6H6 = 16.0 g / 78.11 g/mol

Now, we need to use the stoichiometric coefficients from the balanced equation to relate the moles of C6H6 to the moles of heat released.

The balanced equation shows that the coefficient of C6H6 is 2, which means that for every 2 moles of C6H6 combusted, ΔH° is released. Therefore, we can set up a proportion to find the moles of heat released:

2 moles C6H6 / ΔH° = moles C6H6 / x

Rearranging the equation to solve for x (moles of heat released), we have:

x = (moles C6H6 * ΔH°) / 2

Substituting the values we obtained earlier:

x = (16.0 g / 78.11 g/mol) * -6535 kJ / 2

Finally, we can calculate the moles of heat released and convert it to kilojoules:

moles of heat released = x / 1000

Therefore, the amount of heat released in the combustion of 16.0 g of C6H6 is approximately ________ kJ (calculate the numerical value using the steps above).

To find the amount of heat released in the combustion of 16.0 g of C6H6(l), we need to use the given ΔH° value for the reaction.

First, we calculate the molar mass of C6H6 (benzene):
C = 12.01 g/mol
H = 1.01 g/mol
6C + 6H = 6(12.01 g/mol) + 6(1.01 g/mol) = 78.12 g/mol

Next, we convert 16.0 g of C6H6 to moles:
(16.0 g) / (78.12 g/mol) = 0.205 mol

Since the reaction equation shows that 2 moles of C6H6 react to produce -6535 kJ of heat, we can set up a proportion:

(0.205 mol C6H6) / (2 mol C6H6) = (x kJ) / (-6535 kJ)

Cross-multiplying and solving for x, we get:
x = (0.205 mol C6H6) * (-6535 kJ) / (2 mol C6H6)
x ≈ - 671.02 kJ

Therefore, approximately -671.02 kJ of heat are released in the combustion of 16.0 g of C6H6(l).