1. The value of ΔH° for the reaction below is -6535 kJ.

Question

_________ kJ of heat are released in the combustion of 16.0 g C6H6 (l).

2C6H6 (l) + 15O2 (g) --> 12CO2 (g) + 6H20 (l)

3. Given the following reactions

CaCO3 (s) --> CaO (s) + CO2 (g) ΔH = 178.1 kJ

C (s,graphite) + O2 (g) --> CO2 (g) ΔH = -393.5 kJ

the enthalpy of the reaction

CaCO3 (s) --> CaO (s) + C (s,graphite) + O2 (g)

is ________ kJ.

Answer 1

1. First, we need to find the moles of C6H6 in 16.0 g.

Moles = (16.0 g) / (78.11 g/mol) = 0.2048 moles

Since the given reaction involves 2 moles of C6H6, we need to calculate the amount of heat released per mole of C6H6:

ΔH per mole of C6H6 = (-6535 kJ) / 2 = -3267.5 kJ/mol

Now, we can calculate the heat released during the combustion of 0.2048 moles of C6H6:

Heat released = (0.2048 moles) x (-3267.5 kJ/mol) = -669.0 kJ

So, 669.0 kJ of heat are released in the combustion of 16.0 g C6H6 (l).

3. To find the enthalpy of the desired reaction, we can subtract the ΔH values of the given reactions:

ΔH (desired reaction) = 178.1 kJ - (-393.5 kJ) = 178.1 kJ + 393.5 kJ = 571.6 kJ

So, the enthalpy of the reaction CaCO3 (s) --> CaO (s) + C (s,graphite) + O2 (g) is 571.6 kJ.