Balance the chemical equation using the vector equation approach. if possible, use exact arithmetic or rational format for calculations in balancing the following chemical reaction: pbn6+crmn2o8→pb3o4+cr2o3+mno2+no

15 PbN6 + 44 CrMn2O8 = 5 Pb3O4 + 22 Cr2O3 + 88 MnO2 + 90 NO

Explanation:

Every time you need to balance a chemical equation, you need to end up with the same amount of products that the initial substances.

PbN6+CrMn2O8→Pb3O4+Cr2O3+MnO2+NO

Start analyzing the pairs to balance.

Let's start with the ones to get oxidate (the electrons will be with the products)

The Pb starts alone and as a product, you have Pb 3. You balance this semi equation like this.

3PbN → Pb3O4 + 2e-

With the Cr, the same happens.

2CrMn2O8 → Cr2O3 + 2e-

With N you have to balance the product because you start with N6.

PbN6 → 6NO + 14e-

And then, the one that gets reduced Mn

CrMn2O + 6e- → 2MnO2

Now, combine the oxidate and reduce pairs

3PbN6 → Pb3O4 + 18NO + 44e-

and

2CrMn2O8 + 10e- → Cr2O3 + 4MnO2

Now, equilibrate charges.

3PbN6 → Pb3O4 + 18NO + 44e- + 44H+

2CrMn2O8 + 10e- + 10H+ → Cr2O3 + 4MnO2

and oxygen.

3PbN6 + 22H2O → Pb3O4 + 18NO + 44e- + 44H+

2CrMn2O8 + 10e- + 10H+ → Cr2O3 + 4MnO2 + 5H2O

Now, you can count and see the one that oxidated have to be multiplied by 22 (extra electrons in reduction) and the one reducing by 5 (extra electrons in oxidation)

Then, group and simplify the waters and you'll find that the balanced equation is

15 PbN6 + 44 CrMn2O8 = 5 Pb3O4 + 22 Cr2O3 + 88 MnO2 + 90 NO

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