H2O + XeF6 = HF + XeF4O

water + xenon hexafluoride ⟶ hydrogen fluoride + XeF4O

Balance the chemical equation algebraically: + ⟶ + XeF4O Add stoichiometric coefficients, c_i, to the reactants and products: c_1 + c_2 ⟶ c_3 + c_4 XeF4O Set the number of atoms in the reactants equal to the number of atoms in the products for H, O, F and Xe: H: | 2 c_1 = c_3 O: | c_1 = c_4 F: | 6 c_2 = c_3 + 4 c_4 Xe: | c_2 = c_4 Since the coefficients are relative quantities and underdetermined, choose a coefficient to set arbitrarily. To keep the coefficients small, the arbitrary value is ordinarily one. For instance, set c_1 = 1 and solve the system of equations for the remaining coefficients: c_1 = 1 c_2 = 1 c_3 = 2 c_4 = 1 Substitute the coefficients into the chemical reaction to obtain the balanced equation: Answer: | | + ⟶ 2 + XeF4O

+ ⟶ + XeF4O

water + xenon hexafluoride ⟶ hydrogen fluoride + XeF4O

K_c = ([HF]^2 [XeF4O])/([H2O] [F6Xe1])

rate = -(Δ[H2O])/(Δt) = -(Δ[F6Xe1])/(Δt) = 1/2 (Δ[HF])/(Δt) = (Δ[XeF4O])/(Δt) (assuming constant volume and no accumulation of intermediates or side products)

| water | xenon hexafluoride | hydrogen fluoride | XeF4O formula | | | | XeF4O Hill formula | H_2O | F_6Xe | FH | F4OXe name | water | xenon hexafluoride | hydrogen fluoride | IUPAC name | water | hexafluoroxenon | hydrogen fluoride |

| water | xenon hexafluoride | hydrogen fluoride | XeF4O molar mass | 18.015 g/mol | 245.283 g/mol | 20.006 g/mol | 223.286 g/mol phase | liquid (at STP) | | gas (at STP) | melting point | 0 °C | | -83.36 °C | boiling point | 99.9839 °C | | 19.5 °C | density | 1 g/cm^3 | | 8.18×10^-4 g/cm^3 (at 25 °C) | solubility in water | | | miscible | surface tension | 0.0728 N/m | | | dynamic viscosity | 8.9×10^-4 Pa s (at 25 °C) | | 1.2571×10^-5 Pa s (at 20 °C) | odor | odorless | | |