It took 20.0 mL of NaOH to neutralize 25 mL of 0.50 M HCl. What is the molarity of the base?

5/8 M

Explanation:

NaOH + HCl > NaCl + H2O

in 25 mL of 0.5M acid, we have 1/80 Mol of HCl

it mean we must have 1/80 Mol of NaOH.

20 mL = 1/50 L

(1/80)Mol / (1/50)L = 5/8 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.