A stock solution of Blue #1 has a concentration of 5.736 M. 3 mL of this solution is diluted with 8 mL water. What is the concentration of the resulting solution? Provide your response to 3 significant figures.

1.56 M

Explanation:

This is a dilution process and so a dilution formula is suitably used as follows C1V1 = C2V2 where

C1 = concentration of the stock solution

V1 = volume of the stock solution

C2 = concentration of the resulting (dilute) solution and

V2 = the volume of the resulting (dilute) solution

C1V1 = C2V2 (Making C2 subject of the formula)

C2 = C1V1/V2

Given: C1 = 5.736 M; V1 = 3 Ml; V2 = (3+8) 11 Ml

C2 = 5.736 x 3 / 11

C2 = 1.56 M

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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