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Alka Seltzer Reaction

[tex]\[ \text{NaHCO}_3 + \text{C}_6\text{H}_8\text{O}_7 \rightarrow \text{H}_2\text{O} + \text{CO}_2 + \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 \][/tex]

Sagot :

Sure, let's analyze the given chemical reaction and provide a detailed, step-by-step solution to understanding it.

The provided chemical reaction is:
[tex]$ 3 \text{NaHCO}_3 + \text{C}_6\text{H}_8\text{O}_7 \rightarrow 3 \text{CO}_2 + \text{H}_2\text{O} + \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 $[/tex]

To fully understand the stoichiometry of this reaction, we need to determine the number of moles of each reactant and product involved:
- Reactants:
- Sodium bicarbonate (NaHCO₃)
- Citric acid (C₆H₈O₇)

- Products:
- Carbon dioxide (CO₂)
- Water (H₂O)
- Sodium citrate (Na₃C₆H₅O₇)

### Step-by-Step Breakdown:

1. Balance the Reaction:
The chemical equation provided is already balanced. Each side of the equation has an equal number of each type of atom.

2. Identify the Moles of Each Substance:
We can derive the number of moles of each substance directly from the coefficients in the balanced equation.

- For Sodium bicarbonate (NaHCO₃):
[tex]\[ \text{Coefficient} = 3 \rightarrow 3 \text{ moles of NaHCO}_3 \][/tex]

- For Citric acid (C₆H₈O₇):
[tex]\[ \text{Coefficient} = 1 \rightarrow 1 \text{ mole of C}_6\text{H}_8\text{O}_7 \][/tex]

- For Carbon dioxide (CO₂):
[tex]\[ \text{Coefficient} = 3 \rightarrow 3 \text{ moles of CO}_2 \][/tex]

- For Water (H₂O):
[tex]\[ \text{Coefficient} = 1 \rightarrow 1 \text{ mole of H}_2\text{O} \][/tex]

- For Sodium citrate (Na₃C₆H₅O₇):
[tex]\[ \text{Coefficient} = 1 \rightarrow 1 \text{ mole of Na}_3\text{C}_6\text{H}_5\text{O}_7 \][/tex]

### Final Moles for Each Substance:
- [tex]\( \text{NaHCO}_3 \)[/tex]: 3 moles
- [tex]\( \text{C}_6\text{H}_8\text{O}_7 \)[/tex]: 1 mole
- [tex]\( \text{CO}_2 \)[/tex]: 3 moles
- [tex]\( \text{H}_2\text{O} \)[/tex]: 1 mole
- [tex]\( \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 \)[/tex]: 1 mole

Thus, we can conclude that the number of moles involved in the reaction for each substance are as follows:

[tex]\[ (3, 1, 1, 3, 1) \][/tex]