What happens when NaHCO3 reacts with HCl?

When sodium bicarbonate $({\text{NaHCO}}_3)$ reacts with hydrochloric acid $(\text{HCl})$, it undergoes a chemical reaction that produces carbon dioxide $({\text{CO}}_2)$, water $({\text{H}}_2\text{O})$, and sodium chloride $(\text{NaCl})$. This process is a classic example of an acid-base reaction, where the weak base sodium bicarbonate reacts with the strong acid HCl.

Step-by-Step Explanation of the Reaction

1. Reaction Setup and Balanced Equation

When ${\text{NaHCO}}_3$ comes into contact with $\text{HCl}$, the hydrochloric acid dissociates into hydrogen ions $({\text{H}}^{+})$ and chloride ions $({\text{Cl}}^{-})$. The balanced chemical equation for this reaction is: $${\text{NaHCO}}_3(s)+\text{HCl}(aq)\to \text{NaCl}(aq)+{\text{H}}_2\text{O}(l)+{\text{CO}}_2(g)$$

2. Breaking Down the Reaction Mechanism

The reaction proceeds in two primary stages:

- Step 1: The hydrogen ion $({\text{H}}^{+})$ from $\text{HCl}$ interacts with the bicarbonate ion $({\text{HCO}}_3^{-})$, forming carbonic acid $({\text{H}}_2{\text{CO}}_3)$: $${\text{HCO}}_3^{-}(aq)+{\text{H}}^{+}(aq)\to {\text{H}}_2{\text{CO}}_3(aq)$$

- Step 2: Carbonic acid $({\text{H}}_2{\text{CO}}_3)$ is unstable and quickly decomposes into water $({\text{H}}_2\text{O})$ and carbon dioxide $({\text{CO}}_2)$: $${\text{H}}_2{\text{CO}}_3(aq)\to {\text{H}}_2\text{O}(l)+{\text{CO}}_2(g)$$ Combining these steps gives the complete reaction.

3. Ion Exchange and Product Formation

Sodium ions $({\text{Na}}^{+})$ and chloride ions $({\text{Cl}}^{-})$ combine to form sodium chloride $(\text{NaCl})$, which remains dissolved in the solution: $${\text{Na}}^{+}(aq)+{\text{Cl}}^{-}(aq)\to \text{NaCl}(aq)$$ This results in three distinct products: water, carbon dioxide gas, and sodium chloride.

4. Observations During the Reaction

- Effervescence: The release of ${\text{CO}}_2$ gas causes bubbling or effervescence, a common indicator of this reaction.

- Temperature Change: This reaction is typically exothermic, releasing a small amount of heat as it proceeds.

Calculations: Mole-to-Mole Ratio and Quantitative Analysis

To calculate the amount of each product formed, let's consider an example where 1 mole of ${\text{NaHCO}}_3$ is reacted with an excess of $\text{HCl}$:

1. Mole-to-Mole Ratio: According to the balanced equation: $$1\text{mol}{\text{NaHCO}}_3:1\text{mol}\text{NaCl}:1\text{mol}{\text{H}}_2\text{O}:1\text{mol}{\text{CO}}_2$$

2. Mass Calculation: Given that the molar mass of ${\text{NaHCO}}_3$ is approximately $84.01\text{g/mol}$, reacting 1 mole (or 84.01 g) of ${\text{NaHCO}}_3$ will yield:

- Sodium Chloride (NaCl): $1\text{mol}\times 58.44\text{g/mol}=58.44\text{g}$

- Water (H}_2\text{O): $1\text{mol}\times 18.02\text{g/mol}=18.02\text{g}$

- **Carbon Dioxide (CO}_2\): $1\text{mol}\times 44.01\text{g/mol}=44.01\text{g}$

   Thus, for each mole of ${\text{NaHCO}}_3$ reacted, we expect approximately $58.44\text{g}$ of $\text{NaCl}$, $18.02\text{g}$ of water, and $44.01\text{g}$ of carbon dioxide to be produced.

Summary of Results in Table Format

To summarize the reaction quantitatively:

$$\begin{array}{|ccc|}\hline \text{Compound}& \text{Moles Formed}& \text{Mass (g)}\\ \text{NaCl}& 1\text{mol}& 58.44\text{g}\\ {\text{H}}_2\text{O}& 1\text{mol}& 18.02\text{g}\\ {\text{CO}}_2& 1\text{mol}& 44.01\text{g}\\ \hline\end{array}$$

Key Takeaways

- The reaction between ${\text{NaHCO}}_3$ and $\text{HCl}$ is an acid-base reaction yielding sodium chloride, water, and carbon dioxide.

- The reaction’s main observable indicator is the effervescence due to ${\text{CO}}_2$ gas production.

- The balanced equation and stoichiometric analysis allow for precise calculation of each product formed based on the initial amount of sodium bicarbonate used.