Can someone explain the steps for balancing chemical equations in depth?

explain the steps for balancing chemical equations in depth?

Step 1: The Unbalanced Chemical Equation. The unbalanced chemical equation is given to you.

Step 2: Make a List.

Step 3: Identifying the Atoms in Each Element.

Step 4: Multiplying the Number of Atoms.

Step 5: Placing Coefficients in Front of Molecules.

Step 6: Check Equation.

Step 7: Balanced Chemical Equation.

Steps explained below

Explanation:

To explain balancing of chemical equations, I will make use of an example equation where Hydrogen and oxygen react to form water.

H2 + O2 = H2O

Now, the equation I've listed above is an unbalanced chemical equation. It can be balanced by the following steps;

Step 1: Identify the elements on both the left Hand side and the right hand side.

In this case;

on the left hand side, we have H and O.

On the right hand side, we have H and I also.

Step 2: Identify the number of atoms of each element on both the left and right hand sides.

On the left, H has 2 atoms and O has 2 atoms.

On the right, H has 2 atoms and O has 1 atom.

Step 3: For the equation to be balanced, the number of atoms of each element on the right and left hand side must be the same.

Thus,

O on the left hand side has 2 atoms but on the right hand side it has 1 atom. Thus, we will multiply O on the right by 2 to balance what we have on the left.

So, we now have;

H2 + O2 = 2H2O

Step 4: Check equation: We now have;

H2 + O2 = 2H2O

Our left hand side remains 2 atoms of H and 2 atoms of O. But on the right, we now have;

2 atoms O and 4 atoms of H.

Which means atoms of H is not balanced with the left side.

Step 5: rebalance equation: To rebalance, we multiply H on the left by 2 to give us 2 × 2 = 4 atoms.

Thus, we now have;

2H2 + O2 = 2H2O

Representation of chemical reactivity by symbols follows two absolute rules:

Explanation:

1

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 Mass is conserved.

2

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 Charge is conserved.

What do we mean by this? If we start with 10 g of reactant FROM ALL SOURCES, AT MOST we can get 10 g of product; and in practice we are not even going to get that. Since mass is a fundamental property of atoms and molecules it follows that atoms and molecules are conserved in EVERY chemical reaction.

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Because it is not balanced (why not?), we can reject it out of hand, because we know that it does not reflect reality.

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80 g of reactant gives 80 g of product

...charge and mass are balanced here. And so this is a reasonable representation of reality.

I know I am being absolute when I say that charge and mass are conserved, but this reflects EVERY CHEMICAL REACTION, EVERY EXPERIMENT EVER PERFORMED: when a reaction is studied in detail,  

garbage out has always equalled garbage in.

And how do we know that masses are conserved; that stoichiometry operates? How else but by experiment? How else but by extensive and quantitative study of particular chemical reactions?

Every chemical reaction ever performed (and as far we know, TO BE PERFORMED), displays conservation of mass. And today we have a particle view of chemical reactivity, and our ideas, developed over only some 200-300 years, insist that matter is conserved. That molecules and atoms themselves have discrete masses, which are certainly measurable, supports our notion of conservation of mass.

This idea can be extended to the representation of redox reactions, where we can invoke the electron as a charged particle that is exchanged between species in a redox process. Charge is conserved, as well as mass.

If this all seems a lot to take in, remember you will only be asked to balance equations to which you have already been introduced. Alkanes combust in air to give carbon dioxide and water, iron combines with oxygen to give rust,  

If you can write a chemical equation that balances mass and charge, you have proposed a reasonable chemical pathway. See here for redox equations.

Explanation:

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