Consider the following reaction

2 NO(g) + 2 H2(g) --> N2(g) + 2 H2O(g)

Initially, a mixture of 0.100 M NO, 0.050 M H2 and 0.100 M H2O was allowed to reach equilibrium. There was no N2 present initially. At equilibrium, the concentration of NO was found to be 0.062 M.

How much NO was reacted? M (don't round. should be 3 decimal places)


Answer: .038