Answered

Learners made a mini volcano in a science laboratory by adding sodium bicarbonate to ethanoic acid. They added 100 mL of a [tex]$0.2 \, \text{mol} \, \text{dm}^{-3}$[/tex] ethanoic acid solution to a lump of [tex]$NaHCO_3$[/tex] to start the reaction of the volcano.

The balanced equation for this reaction is:
[tex]\[ CH_3COOH_{(aq)} + NaHCO_{3(s)} \rightarrow CH_3COONa_{(aq)} + H_2O_{(l)} + CO_{2(g)} \][/tex]



Answer :

GinnyAnswer
Let's break down the reaction and solve the problem step-by-step.

### Reaction Details
The reaction given is:
[tex]\[ \text{CH}_3\text{COOH}_{(aq)} + \text{NaHCO}_3_{(s)} \rightarrow \text{CH}_3\text{COONa}_{(aq)} + \text{H}_2\text{O}_{(l)} + \text{CO}_2_{(g)} \][/tex]
This is a reaction between ethanoic acid (acetic acid) and sodium bicarbonate (baking soda), resulting in the formation of sodium acetate, water, and carbon dioxide gas.

### Given:
1. Volume of ethanoic acid solution: 100 ml
2. Concentration of ethanoic acid: 0.2 mol/dm³
3. Mass of sodium bicarbonate (NaHCO₃): 1 g

### Steps to solve:

1. Calculate the moles of ethanoic acid (CH₃COOH) present:
- We know that molarity (M) is given by: [tex]\[ \text{Moles of solute} = \text{Molarity} \times \text{Volume (L)} \][/tex]
- Volume in liters: [tex]\[ 100 \text{ ml} = 0.100 \text{ L} \][/tex]
- Moles of ethanoic acid: [tex]\[ 0.2 \text{ mol} \text{ dm}^{-3} \times 0.100 \text{ L} = 0.020 \text{ moles} \][/tex]

2. Calculate the moles of sodium bicarbonate (NaHCO₃):
- The molar mass of NaHCO₃ (sodium bicarbonate) is approximately 84.01 g/mol.
- Moles of NaHCO₃: [tex]\[ \frac{1 \text{ g}}{84.01 \text{ g/mol}} \approx 0.0119 \text{ moles} \][/tex]

3. Determine the limiting reagent:
- According to the balanced chemical equation, the molar ratio of CH₃COOH to NaHCO₃ is 1:1.
- Hence, the reactant with the smaller number of moles will be the limiting reagent.
- Comparing moles: [tex]\[ \text{Moles of CH₃COOH} = 0.020 \text{ moles} \][/tex]
- [tex]\[ \text{Moles of NaHCO₃} = 0.0119 \text{ moles} \][/tex]
- Since 0.0119 moles of NaHCO₃ are smaller, NaHCO₃ is the limiting reagent.

4. Calculate the moles of CO₂ produced:
- As per the balanced equation, 1 mole of NaHCO₃ produces 1 mole of CO₂ when fully reacted.
- Therefore, the moles of CO₂ produced will be equal to the moles of the limiting reagent (NaHCO₃).
- Moles of CO₂ produced: [tex]\[ 0.0119 \text{ moles} \][/tex]

### Summary of results:
- Moles of ethanoic acid (CH₃COOH): 0.020 moles
- Moles of sodium bicarbonate (NaHCO₃): 0.0119 moles
- Moles of carbon dioxide (CO₂) produced: 0.0119 moles

Thus, from the reaction involving 1 g of sodium bicarbonate and 100 ml of 0.2 mol/dm³ ethanoic acid solution, 0.0119 moles of CO₂ gas are produced.