To determine the mass of the limiting reactant and the excess reactant, we first need to calculate the number of moles of each reactant based on the provided data.
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Calculate the moles of ethanoic acid (CH₃COOH):
- Given concentration of ethanoic acid = 0.2 mol/dm³
- Volume of ethanoic acid = 100 mL = 0.1 dm³
\[ \text{Moles of CH}_3\text{COOH} = \text{Concentration} \times \text{Volume} = 0.2 , \text{mol/dm}^3 \times 0.1 , \text{dm}^3 = 0.02 , \text{moles} \]
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Calculate the moles of sodium bicarbonate (NaHCO₃):
- Given mass of NaHCO₃ = 10 g
- Molar mass of NaHCO₃ = Na (23) + H (1) + C (12) + 3O (16*3) = 23 + 1 + 12 + 48 = 84 g/mol
\[ \text{Moles of NaHCO}_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{10,g}{84,g/mol} \approx 0.119 , \text{moles} \]
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Determine the stoichiometry of the reaction: The balanced equation shows a 1:1 molar ratio between CH₃COOH and NaHCO₃.
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Determine the limiting reactant:
- Moles of CH₃COOH = 0.02 moles
- Moles of NaHCO₃ = 0.119 moles
Since we need 1 mole of NaHCO₃ for every 1 mole of CH₃COOH, we can see that ethanoic acid (CH₃COOH) is the limiting reactant because we have less moles of CH₃COOH compared to the available moles of NaHCO₃.
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Calculate the excess moles of NaHCO₃:
- Moles of NaHCO₃ that reacted = Moles of CH₃COOH = 0.02 moles
- Moles of NaHCO₃ in excess = Moles of NaHCO₃ initial - Moles used = 0.119 - 0.02 = 0.099 moles
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Calculate the mass of excess NaHCO₃: \[ \text{Excess mass of NaHCO}_3 = \text{moles} \times \text{molar mass} = 0.099 , \text{moles} \times 84 , \text{g/mol} \approx 8.316 , g \]
Therefore, the mass of the sodium bicarbonate (NaHCO₃) in excess is approximately 8.32 grams.
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