Give a complete and balanced equation for the reaction between methanoic acid $(\text{HCOOH})$(HCOOH)​ and sodium oxide

The reaction between methanoic acid (HCOOH) and sodium oxide (Na2O) can be written as follows:

2HCOOH + Na2O -> 2HCOONa + H2O

This equation is balanced, meaning there are the same number of each type of atom on both sides of the equation.

Here's the step-by-step process:

1. Write the unbalanced equation: HCOOH + Na2O -> HCOONa + H2O
2. Balance the sodium atoms. There are 2 sodium atoms on the right side and only 1 on the left side, so we put a 2 in front of HCOONa: HCOOH + Na2O -> 2HCOONa + H2O
3. Balance the carbon and oxygen atoms. There are 2 carbon atoms and 4 oxygen atoms on the right side and only 1 carbon atom and 2 oxygen atoms on the left side, so we put a 2 in front of HCOOH: 2HCOOH + Na2O -> 2HCOONa + H2O
4. Check the hydrogen atoms. There are 4 hydrogen atoms on both sides, so the equation is now balanced.