Question

nahco3 + ch3cooh = nach3coo + h2o + co2

3g of NaHCO3 are dissolved into 50mL of deionized water and diluted to 100mL.

5mL of the NaHCO3 solution are pipetted into a reaction flask.

3mL of 0.875M NaHCOOH are placed in a vile in the reaction flask

a) How many moles of NaHCO3 are in the 100mL flask, and how many moles are in the reaction flask.

b) How many moles of CH3COOH are in the reaction flask

c) what is the limiting reagent and how many moles of CO2 evolved?

Answer 1

Ans

a) 3g of NaHCO3 is present in 100mL flask , so the number of moles = 3 / 84 = 0.036 moles

So the number of moles present in 5 mL will be : (0.036 x 5) / 100

= 0.0018 moles

So 0.0018 moles are present in the reaction flask.

b) Number of moles of CH3COOH = 0.875 x 0.003

= 0.002625 moles

c) The reaction utilises equimolar concentration of both the reactants

So the limiting reagent here will be NaHCO3.

Each mole of NaHCO3 produces 1 mole of CO2

So the number of moles of CO2 produced wll be 0.0018 moles.