Let's go through this step by step.
### 1. Definition of the Limiting Reagent
The limiting reagent is the substance in a chemical reaction that is totally consumed first and therefore limits the amount of product formed. Once the limiting reagent is completely used up, the reaction stops and no further products can be formed, even if other reactants are still available in excess.
### 2. Determining the Limiting Reagent
#### Given Data:
- Balanced chemical equation:
[tex]\[ \text{CH}_3\text{COOH} + \text{NaHCO}_3 \rightarrow \text{Na}\text{CH}_3\text{COO} + \text{CO}_2 + \text{H}_2\text{O} \][/tex]
- Molar masses:
- [tex]\(\text{CH}_3\text{COOH}\)[/tex]: 60.05 g/mol
- [tex]\(\text{NaHCO}_3\)[/tex]: 84.01 g/mol
- Volume of CH[tex]\(_3\)[/tex]COOH solution: 2 dm[tex]\(^3\)[/tex]
- Concentration of CH[tex]\(_3\)[/tex]COOH: 0.2 mol/dm[tex]\(^3\)[/tex]
- Mass of NaHCO[tex]\(_3\)[/tex]: 10 g
#### Calculating Moles:
Moles of CH[tex]\(_3\)[/tex]COOH:
The number of moles [tex]\(n\)[/tex] can be calculated via the formula:
[tex]\[ n = \text{volume} \times \text{concentration} \][/tex]
Given:
[tex]\[ \text{volume} = 2 \, \text{dm}^3 \][/tex]
[tex]\[ \text{concentration} = 0.2 \, \text{mol/dm}^3 \][/tex]
So:
[tex]\[ \text{moles of CH}_3\text{COOH} = 2 \, \text{dm}^3 \times 0.2 \, \text{mol/dm}^3 = 0.4 \, \text{mol} \][/tex]
Moles of NaHCO[tex]\(_3\)[/tex]:
The number of moles [tex]\(n\)[/tex] can be calculated via the formula:
[tex]\[ n = \frac{\text{mass}}{\text{molar mass}} \][/tex]
Given:
[tex]\[ \text{mass} = 10 \, \text{g} \][/tex]
[tex]\[ \text{molar mass} = 84.01 \, \text{g/mol} \][/tex]
So:
[tex]\[ \text{moles of NaHCO}_3 = \frac{10 \, \text{g}}{84.01 \, \text{g/mol}} \approx 0.119 \, \text{mol} \][/tex]
#### Determining the Limiting Reagent:
To determine the limiting reagent, compare the moles of each reactant you have calculated:
- Moles of CH[tex]\(_3\)[/tex]COOH: [tex]\(0.4 \, \text{mol}\)[/tex]
- Moles of NaHCO[tex]\(_3\)[/tex]: [tex]\(0.119 \, \text{mol}\)[/tex]
From the balanced chemical equation, the reaction ratio of CH[tex]\(_3\)[/tex]COOH to NaHCO[tex]\(_3\)[/tex] is 1:1. Therefore, the reactant with the fewest moles will be the limiting reagent.
In this case:
[tex]\[ 0.119 \, \text{mol of NaHCO}_3 < 0.4 \, \text{mol of CH}_3\text{COOH} \][/tex]
#### Conclusion:
Since there are fewer moles of NaHCO[tex]\(_3\)[/tex] compared to CH[tex]\(_3\)[/tex]COOH, NaHCO[tex]\(_3\)[/tex] (Sodium bicarbonate) is the limiting reagent in this chemical reaction.
