Question

The reaction of sodium bicarbonate with acetic acid. If you have 0.03M sodium bicarbonate solution (you put 1g into 25 mL of seltzer water) and 0.03M vinegar solution (1.25mL into 25mL of seltzer water) how much carbon dioxide is generated in a 1:1, 2:1 and 1:2 ration of the reactants. Please show work

Answer 1

Ans.

The following assumptions are made –

I. The strength of each reactant is 0.03 M

II. Since the strength of seltzer water is not mentioned it’s assumed that 25.0 mL of the prepared solutions have their strength of 0.03 M to their respective reactants.

III. So, at 1:1 ration, each reactant is taken as 0.03 M, 25.0 mL (= 0.025 L) solution.

IV: At 1:2 or 2:1 ratio reaction mixture is prepared by doubling the volume of required reagent accordingly.

Now,   

Balanced reaction: NaHCO₃ + CH₃COOH ------> CH₃COONa + H₂O + CO₂

Stoichiometry: 1 mol sodium bicarbonate reacts with a mol acetic acid (vinegar) to produce 1 mol CO_2­.

Case 1: NaHCO₃ : CH₃COOH = 1: 1

See stoichiometry, the number of moles of CO₂ produced is equal to the number of moles of either of the reactants if taken in equimolar concentration.

So,

            Moles of NaHCO₃ taken = Molarity x volume (in L) = 0.03 M x 0.025 L = 0.00075 moles

So, number of moles of CO₂ produced = number of moles of reactant taken

                                                            = 0.00075 moles

            Mass of CO2 produced = moles x molar mass

                                                            = 0.00075 moles x (44.0 g mol^-1) = 0.033 g

Case 2: NaHCO₃ : CH₃COOH = 2: 1

Moles of NaHCO₃ in 25.0 mL 0.03 M solution = 0.00075 moles

Moles of CH₃COOH in 25.0 mL 0.03 M solution = 0.00075 moles

At 2:1 ration –

            Moles of NaHCO3 : CH3COOH = 2 : 1 = 2 x (0.00075 moles) : 0.00075 moles

            So, Moles of NaHCO3 : CH3COOH = 0.00150 mol : 0.00075 moles

Since, CH3COOH now is the limiting reactant. So, number of moles of CO₂ produced is equal to the number of limiting reactant.

Hence, moles of CO₂ produced = moles of CH3COOH taken = 0.00075 moles

            Mass of CO2 produced = moles x molar mass

                                                            = 0.00075 moles x (44.0 g mol^-1) = 0.033 g

Case 3: NaHCO₃ : CH₃COOH = 1: 2

Similar to case 2, Since, NaHCO₃ now is the limiting reactant. So, number of moles of CO₂ produced is equal to the number of limiting reactant.

            Hence, moles of CO₂ produced = moles of NaHCO3 taken = 0.00075 moles

Mass of CO2 produced = moles x molar mass

                                                            = 0.00075 moles x (44.0 g mol^-1) = 0.033 g