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

*12 NV + 12 e- → 12 NIV (reduction)*

*6 O-II - 12 e- → 6 O0 (oxidation)*

*Cr(NO3)3* is an **oxidizing** agent, *Cr(NO3)3* is a **reducing** agent.

### Word equation

*chromium nitrate → oxygen + nitrogen dioxide + chromium(III) oxide*

### Input interpretation

*2 CrN3O9 → 3 O2 + 6 NO2 + Cr2O3*

*chromium nitrate oxygen nitrogen dioxide chromium(III) oxide*

### Balanced equation

Balance the chemical equation algebraically:

*2 CrN3O9 → 3 O2 + 6 NO2 + Cr2O3*

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

*c1 2CrN3O9 → c2 3O2 + c36NO2 + c4Cr2O3*

Set the number of atoms in the reactants equal to the number of atoms in the products for ** Cr, N **and

*O:**Cr: c1 = 2c4*

*N: 3c1 = c3*

*O: 9c1 = 2c2 + 2c3 + 3c4*

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, se*t *** c4 = 1** and solve the system of equations for the remaining coefficients:

*c1 = 2 c2= 3/2*

*c3=6 c4=1*

Multiply by the least common denominator, 2, to eliminate fractional coefficients:

*c1 = 4 c2=3*

*c3 = 12 c4 = 2*

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

Answer:

*2 CrN3O9 → 3 O2 + 6 NO2 + Cr2O3*