If 5.300 g the C6H6 is burned and the heat developed from the burning is included to 5691 g that water at 21 °C, what is the final temperature that the water?

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

The ∆H because that the reaction, together written, is -6542 kJ. This to represent the heat produced from burning 2 moles of C6H6(l) and also generating 12 mole of CO2(g), etc.

You are watching: The following equation is the balanced combustion reaction for c6h6:

From the massive of C6H6 given, we can calculate mole of C6H6 combusted. From the value, we deserve to then find the warm that would certainly be generated. ~ finding the warmth generated, us can find the readjust in temperature of the water.

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

heat created = 0.06785 mole x 6542 kJ/2 mole = 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

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