To balance the given chemical equation \(2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2\), we must ensure that the number of atoms of each element is equal on both sides of the equation.
Currently, we have:
- Reactants: 2 Na, 2 H, 2 C, 6 O
- Products: 2 Na, 1 C (from Na2CO3) + 1 C (from CO2) = 2 C, 3 O (from Na2CO3) + 2 O (from CO2) = 5 O
To balance the number of molecules and atoms, we can add \( \text{H}_2\text{O} \) (water) to the products, which would give us additional hydrogen and oxygen.
If we propose adding 1 molecule of water (\( \text{H}_2\text{O} \)), the products become \( \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \).
Now, the counts will look like this:
- Products:
- Sodium: 2 Na (balanced)
- Carbon: 2 C (balanced)
- Hydrogen: 2 H (from \( \text{H}_2\text{O} \), now there are 2 H)
- Oxygen: 5 O (3 from Na2CO3, 2 from CO2, and 1 from \( \text{H}_2\text{O} \) totaling 6, now balanced)
Thus the balanced equation is:
\[ 2 \text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{CO}_2 + \text{H}_2\text{O} \]
Therefore, from the options given, the correct response is:
H2O (Hâ‚‚O).