Question

100.0 mL of 1.00 mol/L NaOH reacted with 55.9 mL of 1.00 mol/L CH3COOH solution and the temperature increased from 20 to 22 C. How many moles of NaOH get and CH3COOH are neutralized?

Answer 1

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

1 mole         1 mole

no of moles of NaOH = molarity * volume in L

                              = 1*0.1 = 0.1 mole

no of moles of CH3COOH =molarity * volume in L

                                     = 1*0.0559 = 0.0559 moles

From balanced equation

1 mole of CH3COOH react with 1 mole of NaOH

0.0559 moles of CH3COOH react with 0.0559 moles of NaOH

Here limiting reagent is CH3COOH

excess reagent is NaOH = 0.1-0.0559 = 0.0441 moles of NaOH not reacted