a lizard drinks water that contains H-2. The H-2 decays and approximately 6.25% remain after 30 seconds. What is the 1/2 life for H-2?

7.5 sec.

Explanation:

A = A₀e⁻^kt => k = ln(A/A₀)/-t => ln(6.25/100)/-30s = 0.0924 s⁻¹

k·t₀.₅=0.693 => t₀.₅=0.693/k = (0.693/0.0924)s = 7.5 s

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.

the answer to your question is c.

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