How many moles are in 250.0 ml of a 0.15 m nitric acid solution?

Explanation:

It is given that molarity is 0.15 M and volume is 250.0 ml. As 1 ml equals 0.001 l.

Hence, 250.0 ml = 0.25 l.

Also, molarity is equal to the number of moles present in liter of solution. Therefore, calculate number of moles as follows.

                   Molarity =

                      0.15 M =

             No. of moles = 0.0375 mol

Thus, we can conclude that moles present in given solution are 0.0375.

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.