Question

what is the reaction of baking soda and vinegar what is the value of the missing mass?

Answer 1

I don’t have the numbers you used, so I can’t give a single numeric answer. Tell me the masses (or the mass of baking soda and the volume and concentration of the vinegar, or the initial and final total mass readings), and I’ll compute it. Meanwhile, here’s how to find the “missing mass” and an example.

1) Reaction and stoichiometry
NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

One mole of sodium bicarbonate reacts with one mole of acetic acid to produce one mole of CO2. If CO2 escapes, the “missing mass” is the mass of CO2 produced.

2) Method (two equivalent ways)
- If you measured initial total mass (reactants) and final mass (after reaction) and gas escaped: missing mass = initial mass − final mass (that lost mass is CO2).
- If you only know amounts of reactants:
a. Convert reactant masses/volumes to moles.
b. Determine the limiting reagent (reaction is 1:1).
c. Moles CO2 = moles of limiting reagent.
d. Mass CO2 = moles CO2 × 44.01 g·mol−1.

Useful molar masses: NaHCO3 = 84.01 g·mol−1, CH3COOH = 60.05 g·mol−1, CO2 = 44.01 g·mol−1.

3) Example
Suppose you used 4.00 g NaHCO3 and 15.0 mL vinegar (assume vinegar is 5.00 g acetic acid per 100 mL, i.e., 5% w/v → 0.75 g CH3COOH in 15.0 mL).

- moles NaHCO3 = 4.00 / 84.01 = 0.0476 mol
- moles CH3COOH = 0.75 / 60.05 = 0.0125 mol
- acetic acid is limiting → moles CO2 = 0.0125 mol
- mass CO2 = 0.0125 × 44.01 = 0.55 g

So the missing mass (mass lost as CO2) would be about 0.55 g in this example.

If you give me your specific masses/volumes (and vinegar concentration if applicable), I’ll calculate the exact missing mass.