A mixture of .10 mol of NO, .050 mol of H2 and .10 mol of H20 is

Question

placed in a 1 liter vessel at 300 K. The following equilibrium is
established.

2 NO + 2 H2 <----> N2 + 2 H20

At equilibrium (N0) = .062 M. What are the equilibrium concentrations
of H2 N2 and H20

Answer 1

2 NO + 2 H2 <----> N2 + 2 H20

initial:
NO = .1
H2 = 0.05
N2 = not given
H2O = 0.1

If NO at equilibrium is 0.062 and it was 0.1 to start, it must have changed by -0.038.

You don't know the N2 at equlibrium because the initial amount is not listed but it is the initial amount + 0.019.
Since the coefficient of H2 is 2 and that of NO is 2, H2O must have decreased by 0.038 to leave 0.05-0.038 = ??

N2 has coefficient of 1; therefore, it has increased by 1/2 NO or H2; thus, it has increased + 0.019.

Same reasoning increase H2O = +0.038 to make equilibrium 0.1 + 0.038 = ??