To determine the mass of the excess substance in the reaction between acetic acid (CH₃COOH) and sodium bicarbonate (NaHCO₃), we first need to balance the chemical equation and understand the stoichiometry involved.
The balanced equation for the reaction is:
\[ \text{CH}_3\text{COOH (aq)} + \text{NaHCO}_3\text{ (s)} \rightarrow \text{CH}_3\text{COONa (aq)} + \text{H}_2\text{O (l)} + \text{CO}_2\text{ (g)} \]
From the balanced equation, we can see that:
- 1 mole of acetic acid reacts with 1 mole of sodium bicarbonate.
Let's denote:
- \( n_a \) = number of moles of acetic acid (CH₃COOH) used.
- \( n_b \) = number of moles of sodium bicarbonate (NaHCO₃) used.
You need to know the initial amounts (in moles or grams) of CH₃COOH and NaHCO₃ to determine which reactant is in excess and how much of the other substance is left over after the reaction.
Step 1: Determine the limiting reactant.
-
Calculate the moles of CH₃COOH and NaHCO₃ you have:
- Molar mass of CH₃COOH = 60.05 g/mol
- Molar mass of NaHCO₃ = 84.01 g/mol
-
Use the following formula to find moles: \[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]
Step 2: Determine the reaction stoichiometry.
- If the moles of CH₃COOH = moles of NaHCO₃, neither is in excess.
- If moles of CH₃COOH > moles of NaHCO₃, then NaHCO₃ is the limiting reactant.
- If moles of NaHCO₃ > moles of CH₃COOH, then CH₃COOH is the limiting reactant.
Step 3: Calculate the mass of the excess reactant.
Based on which reactant is in excess, use the stoichiometry of the equation to calculate how many moles reacted, and how many moles remain of the excess reactant.
Assuming you have \( n_a \) moles of CH₃COOH and \( n_b \) moles of NaHCO₃:
-
If \( n_a > n_b \):
- Moles of CH₃COOH in excess = \( n_a - n_b \)
- Mass of excess CH₃COOH = \( \text{Excess moles} \times 60.05 , \text{g/mol} \)
-
If \( n_b > n_a \):
- Moles of NaHCO₃ in excess = \( n_b - n_a \)
- Mass of excess NaHCO₃ = \( \text{Excess moles} \times 84.01 , \text{g/mol} \)
Example Calculation:
If you start with 100 g of CH₃COOH and 200 g of NaHCO₃:
-
Moles of CH₃COOH: \[ \frac{100 , \text{g}}{60.05 , \text{g/mol}} \approx 1.66 , \text{mol} \]
-
Moles of NaHCO₃: \[ \frac{200 , \text{g}}{84.01 , \text{g/mol}} \approx 2.38 , \text{mol} \]
-
Since \( n_a < n_b \), CH₃COOH is the limiting reagent.
-
Remaining moles of NaHCO₃: \[ 2.38 - 1.66 = 0.72 , \text{mol} \]
-
Mass of excess NaHCO₃: \[ 0.72 , \text{mol} \times 84.01 , \text{g/mol} \approx 60.48 , \text{g} \]
So based on your values, you would determine the excess amount accordingly. Please provide the values of CH₃COOH and NaHCO₃ if you want specific calculations!