A 20 mL volume of a hydrochloric acid solution, HCl (aq), was placed into a beaker containing an appropriate acid-base indicator. It was titrated with a solution of sodium hydroxide, NaOH (aq), until the endpoint was
reached. If 25 mL of sodium hydroxide solution with a concentration of 0.20 mol/L was used, what is the
concentration of the acid?
Select one:
a. 2.5 mol/L
O
b. 0.20 mol/L
C. 0.25 mol/L
d. 2.0 mol/L

C- 0.25mol/L

Explanation:

Hcl+Naoh-Nacl+H2O

Ratio =1:1

0.2moles of Naoh in1000cm3 what about 25=(25×0.2)÷1000=

0.005moles

0.005moles in 20cm³ what about in 1000

(0.005×1000)÷20=0.25mol/L

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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