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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explanation:

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.