balanced combustion reaction for C6H6 Equation: 2C6H6(l)+15O2(g)--> 12CO2+6H2O(l)_6542 KJ

Question

If 5.300 g of C6H6 is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final temperature of the water?

Answer 1

2C6H6(l) + 15O2(g) --> 12CO2(g) + 6H2O(l) ... balanced equation

The ∆H for the reaction, as written, is -6542 kJ. This represents the heat generated from burning 2 moles of C6H6(l) and generating 12 moles of CO2(g), etc.

From the mass of C6H6 given, we can calculate moles of C6H6 combusted. From that value, we can then find the heat that would be generated. After finding the heat generated, we can find the change in temperature of the water.

5.300 g C6H6 x 1 mol/78.11 g = 0.06785 moles

heat generated = 0.06785 moles x 6542 kJ/2 moles = 221.9 kJ

q = mC∆T

221.9 kJ = (5.691 kg)(4.184 kJ/kg/deg)(∆T)

∆T = 221.9 kJ/(5.691 kg)(4.184 kJ/kg/deg)

∆T = 9.3 degrees

Final temperature = 21º + 9.3º = 30.3ºC