Answer
Under these conditions, there are 0.16 mol of $O_2(g)$.
Work Step by Step
1. Write the equilibrium constant expression:
$K = \frac{[Products]}{[Reactants]} = \frac{[H_2]^2[O_2]}{[H_2O]^2}$
** Remember: the coefficients of the reaction are the exponents of the concentrations.
2. Substitute and solve for $[Ó_2]$
$2.4 \times 10^{-3} = \frac{(1.9 \times 10^{-2})^2[O_2]}{(1.1 \times 10^{-1})^2}$
$[O_2] = \frac{2.4 \times 10^{-3}(1.1 \times 10^{-1})^2}{(1.9 \times 10^{-2})^2} = 0.080M$
3. Calculate the number of moles in this 2.0 L container:
$2.0 L \times \frac{0.080mol}{1L} = 0.16mol$
