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### Theory:

The law of mass conservation states that:
The total mass of all the atoms in the reactants should be equal to the total mass of all the atoms in the products; This law is valid or accurate only if the number of atoms of all types of elements on both sides is equal.
Types of chemical balancing methods:
• Trial and error method (direct inspection).
• Fractional method.
• Odd number-even number method.
When balancing a chemical equation keep the following points in mind.
1. Initially, the chemical equation should count the number of times an element appears on both sides of the skeleton equation.
2. The element with the lowest occurrences in both the reactant and product sides must be balanced first. The elements that appear two times, three times, and shortly in an increasing order must all be balanced.
3. When two or more elements appear the same number of times, the metallic element takes precedence over the non-metallic element in balancing. Thus, if there are many metals or non-metals, the metal or non-metal with a higher atomic mass is balanced.
4. The number of reactants and products molecules is expressed as a coefficient.
5. Chemical equation should not change the product formula to make the elements identical.
6. Fractional method of balancing must be employed only for a molecule of an element ($$H_2$$, $$O_3$$, $$P_4$$, ..) not for compound ( $$(H_2O$$, $$NH_3$$,..)
Let's now balance the equation for the hydrogen-oxygen reaction, which produces water.
Example:
Step 1: Write the word equation.
Hydrogen  +  Oxygen  $\to$  Water

Step 2: Write the skeleton equation.
${H}_{2}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}{O}_{2}\phantom{\rule{0.147em}{0ex}}\to {\phantom{\rule{0.147em}{0ex}}H}_{2}O$

Step 3: Choose the first element to balance based on the number of times an element appears on both sides of the skeleton equation.

Step 4: Both elements appear only once in the above case, so the priority is oxygen because it has a greater atomic mass.

Step 5: To balance oxygen, enter $$2$$ before $$H_2O$$ on the right-hand side (RHS).
${H}_{2}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}{O}_{2}\phantom{\rule{0.147em}{0ex}}\to {\phantom{\rule{0.147em}{0ex}}\mathit{2H}}_{2}O$

Step 6: To balance hydrogen, enter $$2$$ near hydrogen $$H_2$$ on the left-hand side (LHS).
${\mathit{2H}}_{2}\phantom{\rule{0.147em}{0ex}}+\phantom{\rule{0.147em}{0ex}}{O}_{2}\phantom{\rule{0.147em}{0ex}}\to {\phantom{\rule{0.147em}{0ex}}\mathit{2H}}_{2}O$
(H= 4  O = 2) (H= 4  O =2)

Now in the above equation, both sides of hydrogen atoms is four and oxygen atoms is two. Hence, the chemical equation is balanced.