[Gweedo8]

A sample of benzene, C6H6, weighing 3.51g was burned in an excess of oxygen in bomb calorimeter. The temperature of the calorimeter rose from 25.00C to 37.18C. If the heat capacity of the calorimeter and contents was 12.05kJ/C, what is the value of q for burning 1.25 mol of benzene at constant volume and 25C? The reaction is
C6H6(l)+15/2O2(g)6CO2(g)+3H2O(l)
Here is my work:

q=c x delta T

12.05x12.18=146.769kg = q calorimeter

So the value of q for burning 1.25 mol of benzene at constant volume and 25C would be
 -146.769

I then converted C6H6 to moles 3.51x78.118=.045mol of C6H6.

Then I multiplied that by 1.25=.05625 moles

Then -146.769kj/.05625=-2609.2266 or -2.61e3kj

is this correct?

[Phlogiston]

A sample of benzene, C6H6, weighing 3.51g was burned in an excess of oxygen in bomb calorimeter. The temperature of the calorimeter rose from 25.00C to 37.18C. If the heat capacity of the calorimeter and contents was 12.05kJ/C, what is the value of q for burning 1.25 mol of benzene at constant volume and 25C? The reaction is
C6H6(l)+15/2O2(g)6CO2(g)+3H2O(l)
Here is my work:

q=c x delta T

12.05x12.18=146.769kg = q calorimeter

So the value of q for burning 1.25 mol of benzene at constant volume and 25C would be
 -146.769

Read the problem very carefully- how much benzene are they referring to for the initial part of the question?

Also, don't forget to put in as many units as you can.  For instance, don't write this:

12.05x12.18=146.769kg = q calorimeter

write this

12.05 kJ/C x 12.18 C = 146.769 kJ = q calorimeter