How many moles of water are produced when 24.3 ml of 0.100 m naoh is mixed with 31.4 ml of 0.08 m acetic acid?

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

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