The balanced combustion reaction for C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ

Question

If  6.200g C6H6 is burned and the heat produced from the burning is added to 5691g of water at 21C, what is the final temperature of the water?

Answer 1

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

I'll assume that the +6542 kJ is per the reaction as shown: 2 moles of C₆H₆ react with 15 moles of O₂ to produce the products shown. We now want to determine the heat proced from 6.200 grams of C₆H₆.

Find the moles C₆H₆ in 6.200 grams by dividing by C₆H₆'s molar mass.

6.200 g/(30.02 g/mole) = 0.2065 moles C₆H₆.

The data supplied suggest that the 6542 kJ is produced by 2 moles of C₆H₆. This amounts to -3271 kJ/mole C₆H₆ . The sign is negative since this is ebnergy that is released by the system.Multiply this times the actual moles to obtain the heat from the reaction:

(0.2065 moles C₆H₆)*(-3271 kJ/mole C₆H₆) = -675.5 kJ

This 675.5 kJ is added to 5691 grams of water, which has a specific heat of 4.186 J/g°C.

We want the temperature change when 675.5 kJ is added to calculate final temperature, T_f .

-675.5 kJ = (4.186 J/g°C)*(5691g)*(21°C - T_f)

T_f = 21.03 °C

Not much change, so please check my calculations and assumptions.

Bob