You figured the q for 2.84g. Now figure it for a mole of ethanol.
Use a proportion.
Here is the question:
A sample of ethanol, C2H5OH, weighing 2.84g was burned in an excess of oxygen in a bomb calorimeter. THe temperature of the calorimeter rose from 24C to 33.73C. If the heat capacity of the calorimeter and contents was 9.63 kj/C. What is the calue of q for burnign 1 mol of ethanol at constant volume and 25.00C?
C2H5OH(l)+3O2(g)-->2CO2(g)+3H2O(l)
Here is my Homework:
q=ms(T final - T initial)
q=2.84x9.63kj/C x (33.73C-25C)= 238.76
But 238.76 is not the correct answer, the book states that it is -00136kJ
Can anyone tell me what I am doing incorrect?
2 answers
2.48/46.068g/mol ethanol=.06165
.06165x9.63kj/C x (33.73C-25C)= 5.183 ??
still not the -00136kJ I am looking for??
.06165x9.63kj/C x (33.73C-25C)= 5.183 ??
still not the -00136kJ I am looking for??
1 answer