An aqueous KNO3 solution is made using 75.1 g of KNO3 diluted to a total solution volume of 1.95 L .Calculate the molarity of the solution. (assume a density of 1.05 g/mL for the solution) Calculate the molality of the solution.
Calculate the mass percent of the solution.

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Explanation:

Hello,

In this case, since the molar mass of potassium nitrate is 101.1 g/mol, we can compute the molarity as follows:

Moreover, as the mass percent is computed as:

Thus, by using the given density of the solution, we obtain:

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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.

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