well, you have 3 mol of H2 for every mol of N2 so your reaction is not limited by either
so
You have the same number of atoms of N2 on left as on right so
.823/.885 = .93 of each reacted
so .07 of each remains
.07 * .885 = .062 Mol of N2 still there
and
.07 * 2.654 = .186 mol of H2 remains
At a certain temperature, 0.885 mol of N2, and 2.654 mol of H2 are placed in a container.
N2+3H2=2NH3
At equilibrium, there is 0.823 mol of NH3 present. Determine the number of mol of N2 and H2 that are present when the reaction is at equilibrium.
Answer in moled of N2 at equilibrium and moles of H2 at equilibrium.
1 answer
1 answer