Question
The kettle will be heated using the combustion of butane, C4H10 (molar mass = 58.12 g/mole).
2C4H10(g) + 13O2(g) à 8CO2(g) + 10H2O(g) DH = – 4773 kJ/mole reaction
How many grams of butane must be burned to provide the energy you calculated in part (c. Energy in part c was 405.48kJ
2C4H10(g) + 13O2(g) à 8CO2(g) + 10H2O(g) DH = – 4773 kJ/mole reaction
How many grams of butane must be burned to provide the energy you calculated in part (c. Energy in part c was 405.48kJ
Answers
DrBob222
dHrxn = 4773 kJ for 2*58.12 grams of butane. You want 405.48 kJ
4773 kJ x (405.48/2*58.12) = ?
4773 kJ x (405.48/2*58.12) = ?