**hint** you only need to add coefficients where the orange boxes are!

Explanation:

Step1: Balance oxygen atoms

The left - hand side has (O_2) (2 oxygen atoms), and the right - hand side has (Na_2O) (1 oxygen atom). To balance oxygen, we can put a coefficient of 2 in front of (Na_2O). So the equation becomes (Na+O_2\rightarrow2Na_2O). Now, the number of oxygen atoms is 2 on both sides.

Step2: Balance sodium atoms

After balancing oxygen, the right - hand side has (2Na_2O), which means there are (2\times2 = 4) sodium atoms. To balance sodium, we put a coefficient of 4 in front of (Na) on the left - hand side. The balanced equation is (4Na + O_2=2Na_2O)

For the table on the left (before balancing):

• For (Na): There is 1 (Na) atom (since the coefficient is 1, implied) on the left - hand side.
• For (O): There are 2 (O) atoms ((O_2)) on the left - hand side.

For the table on the left (after balancing):

• For (Na): There are 4 (Na) atoms (coefficient 4) on the left - hand side.
• For (O): There are 2 (O) atoms ((O_2)) on the left - hand side.

For the table on the right (before balancing):

• For (Na): There are 2 (Na) atoms ((Na_2O)) on the right - hand side.
• For (O): There is 1 (O) atom ((Na_2O)) on the right - hand side.

For the table on the right (after balancing):

• For (Na): There are (2\times2=4) (Na) atoms ((2Na_2O)) on the right - hand side.
• For (O): There are (2\times1 = 2) (O) atoms ((2Na_2O)) on the right - hand side.

Answer:

The coefficients are 4 (for (Na)) and 2 (for (Na_2O)). Left - hand side table (before balancing): (Na:1), (O:2); (after balancing): (Na:4), (O:2) Right - hand side table (before balancing): (Na:2), (O:1); (after balancing): (Na:4), (O:2)