A student titrates 60.0 mL of HNO3(AQ) with 0.30 M NaOH(AQ). Phenolphthalein is used as the indicator.
After adding 42.2 mL of NaOH(AQ), a color change remains for 25 seconds, and the student stops the
titration. Write a balanced equation and calculate the molarity of the of HNO3(AQ), with sig figs and units

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.

2b2t

explanation:

2b2t

it is the most commonly accepted one