20.00mL of 0.55M CH3COOH solution is mixed with 9.95mL of 1.0M NaHCO3 solution.

The reaction occurring is CH3COOH + NaHCO3 --> CH3CHOO + Na + H2O +CO2.

Which is the limiting reagent in this reaction?

The next question is, how may grams of CO2 will be produced as a result?

I work these limiting reagent (LR) problems the long way.

mols CH3COOH = M x L = ?
mols NaHCO3 = M x L = ?

Using the coefficients in the balanced equation, convert mols CH3COOH to mols CO2.
Do the same and convert mols NaHCO3 to mols CO2.
It is likely these two numbers will not be the same; the correct value in LR problems is ALWAYS the smaller value and the reagent responsible for that number is the LR.
For the next problem, convert to mols CO2 using the smller number of mols CO2.
grams CO2 = mols CO2 x molar mass CO2.

To determine the limiting reagent in a chemical reaction, you need to compare the moles of each reactant present and their stoichiometric ratio in the balanced equation.

First, let's calculate the number of moles for each reactant:

For CH3COOH:
Volume (V) = 20.00 mL = 0.02000 L
Molarity (M) = 0.55 M

Number of moles = V * M = 0.02000 L * 0.55 mol/L = 0.011 mol

For NaHCO3:
Volume (V) = 9.95 mL = 0.00995 L
Molarity (M) = 1.0 M

Number of moles = V * M = 0.00995 L * 1.0 mol/L = 0.00995 mol

Now, let's compare the moles of the two reactants. According to the balanced equation, the stoichiometric ratio between CH3COOH and NaHCO3 is 1:1. This means that for every 1 mole of CH3COOH reacted, 1 mole of NaHCO3 will also react.

In this case, we have 0.011 moles of CH3COOH and 0.00995 moles of NaHCO3. Since the moles of NaHCO3 are less than the moles of CH3COOH, NaHCO3 is the limiting reagent.