Amixture of 0.10 mol of no, 0.050 mol of h2, and 0.10 mol of h2o is placed in a 1.0-l vessel at 300 k. the following equilibrium is established: 2no(g)+2h2(g)⥫⥬==n2(g)+2h2o(g) at equilibrium [no]=0.062m. calculate the equilibrium concentrations of h2, n2, and h2o.
calculate the equilibrium concentration of n2

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

Explanation:

The strategy here is to account for the species at equilibrium given that the concentration of [NO]=0.062M at equilibrium is known and the quantities initially present and its stoichiometry.

                  2NO(g)         +    2H2(g)    ⇒        N2(g)      +         2H2O(g)

i  mol            0.10                   0.050                                             0.10

c mol            -0.038                -0.038                +0019                +0.038                                                

e mol            0.062                 0.012                  00.019               0.057

Since the volume of the vessel is 1.0 L, the concentrations in molarity are:

[NO] =   0.062 M

[H2] =    0.012 M

[N2] =    0.019 M

[H2O] =  0.057 M

answer: find it ur self

explanation:

if you mean hydrogen and oxygen then water is the product