Calculate ∆h0 for the reaction 2 n2(g) + 5 o2(g) −→ 2 n2o5(g) given the data h2(g) + 1 2 o2(g) −→ h2o(ℓ) ∆h 0 f = −283.7 kj/mol n2o5(g) + h2o(ℓ) −→ 2 hno3(ℓ) ∆h0 = −78.5 kj/mol 1 2 n2(g) + 3 2 o2(g) + 1 2 h2(g) −→ hno3(ℓ) ∆h0 f = −173 kj/mol answer in units of kj.

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.