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

To balance the reaction between methanoic acid ($\text{HCOOH}$) and sodium ($\text{Na}$), we start by identifying the products formed from the reaction. When methanoic acid reacts with sodium, it typically produces sodium formate ($\text{HCOONa}$) and hydrogen gas ($\text{H}_2$).

Balanced Reaction Equation

1. Identify Reactants and Products:

   • Reactants: Methanoic acid ($\text{HCOOH}$), Sodium ($\text{Na}$)
   • Products: Sodium formate ($\text{HCOONa}$), Hydrogen gas ($\text{H}_2$)

2. Write the Unbalanced Equation: $\text{HCOOH} + \text{Na} \rightarrow \text{HCOONa} + \text{H}_2$

3. Balance the Equation:

   • We have 1 methanoic acid and 1 sodium producing 1 sodium formate and 1 mole of hydrogen gas.
   • The equation is already balanced; however, we need to ensure sodium is balanced since it produces two moles of hydrogen.

Finalizing, we find that to balance the number of sodium atoms and hydrogen atoms, we require 2 moles of sodium:

$\text{HCOOH} + 2 \text{Na} \rightarrow \text{HCOONa} + \text{H}_2 + \text{Na}$

Final Balanced Equation:

The complete and balanced equation for the reaction is: $\text{HCOOH} + \text{Na} \rightarrow \text{HCOONa} + \frac{1}{2} \text{H}_2$

So the complete balanced reaction is: $\text{HCOOH} + 2 \text{Na} \rightarrow 2 \text{HCOONa} + \text{H}_2$ This represents the reaction where 1 mole of methanoic acid reacts with 2 moles of sodium to produce 2 moles of sodium formate and 1 mole of hydrogen gas.