### This is an **oxidation-reduction** (redox) reaction:

* OII + 2 e- → O0 (reduction)*

*OII + 2 e- → O0 (reduction)*

* O-II - 2 e- → O0 (oxidation)*

*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:

**Answer**:

* F2O + H2O → HF + O2*