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

Question

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Solution

To balance the reaction between methanoic acid (HCOOH\text{HCOOH}) and sodium (Na\text{Na}), we start by identifying the products formed from the reaction. When methanoic acid reacts with sodium, it typically produces sodium formate (HCOONa\text{HCOONa}) and hydrogen gas (H2\text{H}_2).

Balanced Reaction Equation

  1. Identify Reactants and Products:

    • Reactants: Methanoic acid (HCOOH\text{HCOOH}), Sodium (Na\text{Na})
    • Products: Sodium formate (HCOONa\text{HCOONa}), Hydrogen gas (H2\text{H}_2)
  2. Write the Unbalanced Equation: HCOOH+NaHCOONa+H2 \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:

HCOOH+2NaHCOONa+H2+Na \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: HCOOH+NaHCOONa+12H2 \text{HCOOH} + \text{Na} \rightarrow \text{HCOONa} + \frac{1}{2} \text{H}_2

So the complete balanced reaction is: HCOOH+2Na2HCOONa+H2 \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.

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