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
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 CO2.
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
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