Asked by shankar
the combustion of co2 gives -393.5kj/mol calculate the heat of formation of 35.2grams of co2
Answers
Answered by
Mia
enthalpy of combustion of CO2=-393.5 kJ/mol
enthalpy of formation of CO2 = -enthalpy of combustion of CO2 = -(-393.5 kJ/mol) = 393.5 kJ/mol
number of moles of CO2 = mass CO2/molecular mass of CO2 = 35.2 g/ (12.01+16.0x2)g/mol =0.800 mol
1 mol of CO2 formed using 393.5 kJ
0.800 mol of CO2 formed using 0.8x393.5 kJ = 314.8 kJ
enthalpy of formation of CO2 = -enthalpy of combustion of CO2 = -(-393.5 kJ/mol) = 393.5 kJ/mol
number of moles of CO2 = mass CO2/molecular mass of CO2 = 35.2 g/ (12.01+16.0x2)g/mol =0.800 mol
1 mol of CO2 formed using 393.5 kJ
0.800 mol of CO2 formed using 0.8x393.5 kJ = 314.8 kJ
Answered by
DrBob222
How does one combust CO2? Burn it with oxygen? fat chance.
Answered by
max
Thanks this answer is absolutely correct
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.