posted by Scott L. .
Consider the reaction below:
4NH3(g) + 3O2(g) -> 2N2 + 6H20, K=10^80 @ certain temp
initially, all the reactants and products have concentrations equal to 12M. At equilibrium, what is the approximate concentration of oxygen?
k=10^80 is a big number, so this is a forward reaction (very close to completion). So figure the limiting reactant: Looks like NH3 goes first, so when the 12M*Volume is used up, one will have used 3/4*12*volume of O2, which is 9 liters, leaving 3 liters, so 3M is the final oxygen concentration
Oh ok I understand now. Thanks!