You have 1.5L of a 14M stock solution of nitric acid. You need to make 3L of a 5M solution of nitric acid.

see explanation

Explanation:

(Molarity x Volume)concentrate = (Molarity x Volume)diluted

14M x V(concentrate) = 5M x 3L

V(concentrate) = (5M x 3L)/14M = 1.07L of concentrate needed

Mixing:

On mixing water directly into nitric acid at high concentrations the solution tends to hydrate very rapidly and release heat extensively and perhaps splash back solution onto surroundings. Thus, one should apply caution and wear safety clothing; i.e., gloves and protective eye ware while working with such systems.

With the hazards of mixing in mind, one should first add a small quantity of water solvent into the mixing vessel, perhaps 0.50L - 1.00L of the solvent water. Then slowly add the needed amount (in this case 1.07L of the 14M nitric acid solution) of nitric acid into the solvent water. Then dilute further with solvent water up to but not to exceed the 3L reference mark of the mixing container. Mix and carefully transfer the new solution (3L of 5M HNO₃) into an appropriate storage container and seal.

Note: As with any chemical of interest, one should review the MSDS for the reagent being used, noting hazards and proper laboratory handling in order to avoid accidents and/or injury that may occur as a result of improper use.

As a general rule, always add acid into water, NEVER water into acid. Remember A&W (acid into water) root beer.

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.