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This is an oxidation-reduction (redox) reaction:

OII + 2 e- → O0 (reduction)

O-II - 2 e- → O0 (oxidation)

F2O is an oxidizing agent, H2O is a reducing agent (synproportionation (comproportionation)).

Word equation

oxygen difluoride + water → hydrogen fluoride + oxygen

Input interpretation

F2O + H2O → HF + O2

Oxygen difluoride water hydrogen fluoride oxygen

Balanced equation

Balance the chemical equation algebraically:

F2O + H2O → HF + O2

Add stoichiometric coefficients, c,, to the reactants and products:

c1F2O + c2H2O → c3 HF + c4O2

Set the number of atoms in the reactants equal to the number of atoms in the products for F, O and H:

F: 2c1 = c3

O: c1 + c2 = 2c4

H: 2c2 = c3

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 c1 = 1 and solve the system of equations for the remaining coefficients:

c1 = 1 c2=1

c3 = 2 c4=1

Substitute the coefficients into the chemical reaction to obtain the balanced equation:


F2O + H2O → HF + O2