Determine the energy value for the combustion of butane, using the bond energies from the table. Respond in terms of the energy value (not the enthalpy change, the sign is important here). Answer to 3 significant figures in units of kJ/mol. C 4H 10 + 6.5O 2 → 4CO 2 + 5H 2O

Bond Bond Energy (kJ/mol)
H–H 432
O=O 494
O–H 460
C–H 410
C–O 360
C=O 799
C≡O 1071
C–C 347
C=C 611
C=C, aromatic 519
N=O 623
N≡N 941

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

i think its 100

i hope this !