0.002512 moles of H2O
The reaction between acetic acid ( CH3COOH) and NaOH is given by the equation;
CH3COOH + NaOH ------> CH3COONa + H2O
Number of moles of CH3COOH = molarity × volume in litres
= 0.08 × 31.4/1000
= 2.512 × 10^-3
Similarly number of moles of NaOH = 1 × 24.3/1000
= 0.0243
From the reaction the mole ratio of CH3COOH : NaOH
Therefore; 0.0243 moles of NaOH will react with 0.0243 moles of CH3COOH but no.of moles of CH3COOH given in the question are 0.002512 moles, which is less than what is required.
Thus; CH3COOH is the limiting reagent and amount of products produced will depend on amount of CH3COOH only.
Since; 1 mole of CH3COOH gives 1 mole of water.
Then; 0.002512 moles of CH3COOH will give 0.002512 moles of H2O
The number of moles produced from the reaction of 24.3 mL of 0.100 M NaOH with 31.4 mL of 0.08M acetic acid is 2.43x10⁻³ moles.
The acid-base reaction between NaOH and CH₃COOH is the following:
NaOH + CH₃COOH ⇄ H₂O + CH₃COONa
To find the number of moles of water produced we need to calculate the number of moles of NaOH and CH₃COOH
[tex] n = C*V [/tex]
Where:
n: is the number of moles
C: is the concentration
V: is the volume
So, the number of moles of NaOH and CH₃COOH is:
[tex] n_{NaOH_{i}} = C_{NaOH}*V_{NaOH} = 0.100 M*0.0243 L = 2.43 \cdot 10^{-3} moles [/tex]
[tex] n_{{CH_{3}COOH}_{i}} = C_{CH_{3}COOH}*V_{CH_{3}COOH} = 0.08 M*0.0314 L = 2.51 \cdot 10^{-3} moles [/tex]
Now, we need to find the limiting reactant knowing that 1 mol of NaOH reacts with 1 mol of CH₃COOH:
[tex] n_{NaOH} = \frac{1\: mol NaOH}{1\: mol CH_{3}COOH}*2.51 \cdot 10^{-3} \: moles\: CH_{3}COOH = 2.51 \cdot 10^{-3} \: moles [/tex]
We can see that we need 2.51x10⁻³ moles of NaOH to react with CH₃COOH, and initially, we have 2.43x10⁻³ moles of NaOH, so the limiting reactant is NaOH.
Knowing that 1 mol of NaOH reacts with 1 mol of CH₃COOH to produce 1 mol of H₂O and 1 mol of CH₃COONa, the number of moles of water produced is:
[tex] n_{H_{2}O} = n_{NaOH} = 2.43 \cdot 10^{-3} moles [/tex]
Therefore, the number of moles of water produced is 2.43x10⁻³ moles.
You can find more about the acid-base reaction here:
I hope it helps you!
