To calculate the volume of carbon dioxide (\( CO_2 \)) produced at standard temperature and pressure (STP), we first need to determine the number of moles of the reactants involved and the stoichiometry of the reaction.
Step 1: Calculate moles of ethanoic acid (\( CH_3COOH \))
We have 100 mL (0.1 L) of a 0.2 mol/dm³ solution of ethanoic acid:
\[ \text{Moles of } CH_3COOH = \text{concentration} \times \text{volume} = 0.2 , \text{mol/dm}^3 \times 0.1 , \text{dm}^3 = 0.02 , \text{mol} \]
Step 2: Calculate moles of sodium bicarbonate (\( NaHCO_3 \))
The molar mass of \( NaHCO_3 \) is calculated as follows:
- \( Na = 23 , \text{g/mol} \)
- \( H = 1 , \text{g/mol} \)
- \( C = 12 , \text{g/mol} \)
- \( O_3 = 16 , \text{g/mol} \times 3 = 48 , \text{g/mol} \)
Total molar mass of \( NaHCO_3 \): \[ 23 + 1 + 12 + 48 = 84 , \text{g/mol} \]
Now, we convert 10 g of \( NaHCO_3 \) to moles: \[ \text{Moles of } NaHCO_3 = \frac{10 , \text{g}}{84 , \text{g/mol}} \approx 0.119 , \text{mol} \]
Step 3: Determine the limiting reagent
The balanced equation is: \[ CH_3COOH(aq) + NaHCO_3(s) \rightarrow CH_3COONa(aq) + H_2O(l) + CO_2(g) \]
From the equation, 1 mole of \( CH_3COOH \) reacts with 1 mole of \( NaHCO_3 \).
- We have 0.02 moles of \( CH_3COOH \).
- We have 0.119 moles of \( NaHCO_3 \).
Since we have less \( CH_3COOH \) compared to \( NaHCO_3 \), \( CH_3COOH \) is the limiting reagent.
Step 4: Calculate the amount of \( CO_2 \) produced
According to the balanced equation, 1 mole of \( CH_3COOH \) produces 1 mole of \( CO_2 \). Therefore, if we start with 0.02 moles of \( CH_3COOH \), it will produce 0.02 moles of \( CO_2 \).
Step 5: Calculate the volume of \( CO_2 \) at STP
At STP (Standard Temperature and Pressure), 1 mole of an ideal gas occupies 22.4 liters.
\[ \text{Volume of } CO_2 = \text{moles of } CO_2 \times 22.4 , \text{L/mol} = 0.02 , \text{mol} \times 22.4 , \text{L/mol} = 0.448 , \text{L} \]
Conclusion
The volume of \( CO_2 \) produced at STP is 0.448 liters (or 448 mL).