Select from the unit conversion factor options below to determine the mass in grams of 7.00 × 1022 molecules of NaOH. 1 mol NaOH / 6.02 x 1023 molecules NaOH, 6.02 x 1023 molecules NaOH / 1 mol NaOH
40 g NaOH / 1 mol, 1 mol / 40 g NaOH4

a
6.65 g
b
7.0 g
c
4.65 g
d
46.5 g

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.