Question

The following equation is the balanced combustion reaction for C6H6:

2C₆H₆(l) + 15O₂(g) ------> 12CO₂(g)+6H₂O(l)+6542 kJ

If 6.900 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?

please answer this is due today thank you

Answer 1

2C6H6(l) + 15O2(g) -------> 12CO2(g) + 6H2O(l) + 6542kJ

So, 6542 KJ of heat is relesed by two mole of C6H6 Combustion

Therefore,

Heat released per 1 mole of combustion of C6H6 = 3271kJ = 3271000J

Mass of C6H6 = 6.900g

Molar mass of C6H6 = 78.11g/mol

No of mole of C6H6 = 6.900g/78.11(g/mol)=0.08834

Therefore,

Heat released by 0.08834 mole of C6H6 burning = 0.08834 × 3271000J = 288960J

Heat released by reaction = Heat absorbed by water

q=m × ∆T × C

where,

q = Heat absorbed by water,

m = mass of water ,5691g

∆T = Temperature raise

C = heat capacity of water , 4.184J/g ℃

Therefore,

288960J = 5691g × ∆T × 4.184J/g ℃

∆T = 12.14℃

Initial temperature =21℃

Therefore,

Final temperature = 21℃ + 12.14℃ = 33.14℃