To solve this problem, we need to follow a series of systematic steps:
### Step 1: Understand the chemical equation
The chemical reaction between sodium bicarbonate (NaHCO₃) and acetic acid (CH₃COOH) is given by the balanced chemical equation:
[tex]\[ \text{NaHCO}_3 + \text{CH}_3\text{COOH} \longrightarrow \text{CO}_2 + \text{H}_2\text{O} + \text{CH}_3\text{COONa} \][/tex]
This shows a 1:1 molar ratio between sodium bicarbonate and acetic acid, meaning each mole of acetic acid reacts with one mole of sodium bicarbonate.
### Step 2: Identify the given information
We are given that:
- The amount of acetic acid we have is 6.601 moles.
### Step 3: Determine the moles of sodium bicarbonate needed
Since the reaction is in a 1:1 molar ratio, the moles of sodium bicarbonate required will be equal to the moles of acetic acid. Therefore, we need 6.601 moles of sodium bicarbonate.
### Step 4: Find the molar mass of sodium bicarbonate
The molar mass of sodium bicarbonate (NaHCO₃) is calculated as follows:
- Sodium (Na): Approximately 23 g/mol
- Hydrogen (H): Approximately 1 g/mol
- Carbon (C): Approximately 12 g/mol
- Oxygen (O): Approximately 16 g/mol
So:
[tex]\[ \text{Molar mass of NaHCO}_3 = 23 + 1 + 12 + (3 \times 16) = 23 + 1 + 12 + 48 = 84 \, \text{g/mol} \][/tex]
### Step 5: Calculate the grams of sodium bicarbonate needed
To find out how many grams of sodium bicarbonate are required, we use the formula:
[tex]\[ \text{Grams of NaHCO}_3 = \text{Moles of NaHCO}_3 \times \text{Molar mass of NaHCO}_3 \][/tex]
Given:
- Moles of NaHCO₃ needed: 6.601 moles
- Molar mass of NaHCO₃: 84 g/mol
Calculating:
[tex]\[ \text{Grams of NaHCO}_3 = 6.601 \, \text{moles} \times 84 \, \text{g/mol} = 554.484 \, \text{g} \][/tex]
### Conclusion
Therefore, the mass of sodium bicarbonate required to react completely with 6.601 moles of acetic acid is 554.484 grams.
