A certain amount of H2S was added to a 2.0 L flask and allowed to come to equilibrium. At equilibrium, 0.072 mol of H2 was found. How many moles of H2S were originally added to the flask?

0.098 moles H₂S

Explanation:

The reaction that takes place is

2H₂ (g) + S₂ (g) ⇄ 2H₂S (g) keq = 7.5

We can express the equilibrium constant as:

keq = [H₂S]² / [S₂] [H₂]² = 7.5

With the volume we can calculate the equilibrium concentration of H₂:

[H₂] = 0.072 mol / 2.0 L = 0.036 M

The stoichiometric ratio tells us that the concentration of S₂ is half of the concentration of H₂:

[S₂] = [H₂] / 2 = 0.036 M / 2 = 0.018 M

Now we can calculate [H₂S]:

7.5 = [H₂S]² / (0.018*0.036²) [H₂S] = 0.013 M

So 0.013 M is the concentration of H₂S at equilibrium.

This would amount to (0.013 M * 2.0 L) 0.026 moles of H₂S The moles of H₂ at equilibrium are equal to the moles of H₂S that reacted.

Initial moles of H₂S - Moles of H₂S that reacted into H₂ = Moles of H₂S at equilibrium

Initial moles of H₂S - 0.072 mol = 0.026 mol

Initial moles of H₂S = 0.098 moles H₂S