Question

how many moles of water are produced when 24.3 mL of .100 M NaOH is mixed with 22.4 mL of .08 M acetic acid

Answer 1

Number of moles of NaOH , n = Molarity x volume in L

                                           = 0.100 M x 24.3 mL x 10^-3 L/mL

                                           = 2.43x10^-3 mol

Number of moles of CH3COOH , n = Molarity x volume in L

                                                  = 0.08 M x 22.4 mL x 10^-3 L/mL

                                             = 1.79x10^-3 mol

NaOH + CH3COOH -----> CH3COONa + H2O

According the above balanced equation,

1 mole of NaOH reacts with 1 mole of CH3COOH

1.79x10^-3 mol of NaOH reacts with 1.79x10^-3 mol of CH3COOH

So (2.43x10^-3 mol - 1.79x10^-3 mol ) of NaOH left unreated so NaOH is the excess reactant.

Since all the mass of CH3COOH is completly reacted it is the limiting reactant.

From the reaction ,

1 mole of CH3COOH produces 1 mole of water

1.79x10^-3 mol of CH3COOH produces 1.79x10^-3 mol of water

Therefore the number of moles of water produced is 1.79x10^-3 mol