Calculate the equilibrium concentrations of H2O(g), Cl2(g), HCl(g), and O2(g) at 298 K if the initial concentrations are [H2O] = 0.050 M and [Cl2] = 0.100 M. The equilibrium constant Kc for the reaction H2O(g) + Cl2(g) ⇌ 2 HCl(g) + O2(g) is 8.96 × 10^-9 at 298 K.

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Write the balanced chemical equation for the reaction: H2O(g) + Cl2(g) ⇌ 2 HCl(g) + O2(g).

Set up an ICE (Initial, Change, Equilibrium) table to track the concentrations of each species. Initially, [H2O] = 0.050 M, [Cl2] = 0.100 M, and [HCl] and [O2] are 0 M.

Define the change in concentration for the reactants and products in terms of x, where x is the amount of H2O and Cl2 that reacts. The changes will be: -x for H2O and Cl2, +2x for HCl, and +x for O2.

Express the equilibrium concentrations in terms of x: [H2O] = 0.050 - x, [Cl2] = 0.100 - x, [HCl] = 2x, and [O2] = x.

Substitute these expressions into the equilibrium constant expression Kc = ([HCl]^2[O2])/([H2O][Cl2]) and solve for x using the given Kc value of 8.96 × 10^-9.

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