Question

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 \(Na+O_2
ightarrow2Na_2O\). 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): \(Na:4\), \(O:2\)