Question

For the formation of ammonia, the equilibrium constant is known to be 5.2 × 10^−5 at 25◦C. After analysis, it is determined that [N2] =3M and [H2]=0.8M, both at equilibrium. How many grams of ammonia are in the 10L reaction vessel at equilibium? Use the equilibrium equation
N2(g) + 3H2(g) ⇀ ↽ 2NH3(g)
Answer in units of g

Answer 1

K = (NH3)^2/(N2)(H2)^3
You know N2 and H2, substitute with Kc and find NH3 in M = mols/L.
Then to find mols in 10L, that is
mols = M x L
Then mol = grams/molar mass. You know molar mass and mols, solve for grams.

Answer 2

1.2 g NH3