4 NH3(g) + 3 02(g) → 2 N2(g) + 6 H2O(g) If the standard molar heats of formation of ammonia, NH3(g), and gaseous water, H20(g), are -46
kJ/mol and -242 kJ/mol, respectively, what is the value of AH 298 for the reaction represented
above?

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.

some sources quote the lethal dose, for an average 170lbs opiate-naive individual is between 75 and 375 mg

your answer would be “dry”

i'm pretty sure it is letter c.) car