Question

Consider the equation: S + 3O2 and SO3. Is this equation balanced? why or why not​

Answer 1

Answer:

This equation is not balanced because you don't have the same amount of each element on each side of the chemical reaction. The balanced equation is:

2 S + 3 $O_{2}$ ⇒ 2 $SO_{3}$

Explanation:

The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.

Then, you must balance the chemical equation. For that, you must first look at the subscripts next to each atom to find the number of atoms in the equation. If the same atom appears in more than one molecule, you must add its amounts.  

The coefficients located in front of each molecule indicate the amount of each molecule for the reaction. This coefficient can be modified to balance the equation, just as you should never alter the subscripts.

By multiplying the coefficient mentioned by the subscript, you get the amount of each element present in the reaction.

Then, taking into account all of the above, you can determine the amount of elements on each side of the equation:

Left side: 1 sulfur S and 6 oxygen O (coefficient 3 multiplied by sub-index 2)

Right side: 1 sulfur S and 3 oxygen O (subindice value)

As you can see, you have the same amount of sulfur on both sides of the equation but the amount of oxygen is different. This indicates that the chemical equation is not balanced. To balance it, as the amount of sulfur is the same, the amount of oxygen must be balanced, which is different on each side of the reaction.

A simple way is to balance the equation is to multiply the product by 2, that is, add a coefficient 2 in front of the SO3 molecule, the reaction being as follows:

S + 3 $O_{2}$ ⇒ 2 $SO_{3}$

Now the amount of elements on each side of the equation is:

Left side: 1 sulfur S and 6 oxygen O (coefficient 3 multiplied by subindice 2)

Right side: 2 sulfur S and 6 oxygen O (coefficient 2 multiplied by subindice 3)

The oxygen is now balanced, but the amount of sulfur on both sides of the reaction varies. To balance the quantities of sulfur, as now on the right side you have an amount of 2, you can add the coefficient to sulfur. The chemical equation is as follows:

2 S + 3 $O_{2}$ ⇒ 2 $SO_{3}$

Now the amount of elements on each side of the equation is:

Left side: 2 sulfur S and 6 oxygen O (coefficient 3 multiplied by subindice 2)

Right side: 2 sulfur S and 6 oxygen O (coefficient 2 multiplied by subindice 3)

Finally you have the same amount of sulfur and oxygen on both sides of the reaction. So the chemical equation is finally balanced.

Answer 2

The balanced equation is
2SO
2
+
O
2
→
2SO
3