This is an oxidation-reduction (redox) reaction:

10 F0 + 10 e- → 10 F-I (reduction)

2 N-III - 10 e- → 2 NII (oxidation)

F2 is an oxidizing agent, NH3 is a reducing agent

.

### Word equation

fluorine + ammonia → tetrafluorohydrazine + hydrogen fluoride

### Input interpretation

F2 + NH3 → F4N2 + HF

Fluorine ammonia tetrafluorohydrazine hydrogen fluoride

### Balanced equation

Balance the chemical equation algebraically:

F2 + NH3 → F4N2 + HF

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

c1F2 + c2NH3 → c3F4N2 + c4HF

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

F: 2c1 = 4c3 + c4

H: 3c2 = c4

N: c2=2c3

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 c3 = 1 and solve the system of equations for the remaining coefficients:

c1=5 c2=2

c3=1 c4=6

Substitute the coefficients into the chemical reaction to obtain the balanced

equation: