Question

what is the reaction of baking soda and vinegar what is the value of the missing mass (g) for the product?

Answer 1

The reaction (baking soda = sodium bicarbonate, NaHCO3; vinegar ≈ acetic acid, CH3COOH) is:

NaHCO3 + CH3COOH → CH3COONa + H2O + CO2

To find a “missing mass” you need to say which mass you want (mass of a particular product, mass of CO2 released, mass remaining in the container if CO2 escapes) and give the masses of the starting materials (or their amounts in moles). The general steps:

1. Write the balanced equation (above).
2. Convert given masses to moles (moles = mass / molar mass).
- Molar masses: NaHCO3 = 84.01 g/mol; CH3COOH = 60.05 g/mol; CH3COONa = 82.03 g/mol; H2O = 18.02 g/mol; CO2 = 44.01 g/mol.
3. Determine the limiting reagent (compare mole ratio; the reaction is 1:1).
4. Use the limiting reagent’s moles to find moles (and then mass) of each product (1 mol NaHCO3 → 1 mol of each product).
5. If CO2 escapes, subtract its mass from the total product mass to get the mass remaining in the container.

Worked example (if 5.00 g NaHCO3 reacts with 5.00 g CH3COOH, assuming pure acetic acid):
- moles NaHCO3 = 5.00 / 84.01 = 0.05952 mol (limiting reagent)
- moles CH3COOH consumed = 0.05952 mol → mass consumed = 0.05952 × 60.05 = 3.572 g (so 1.428 g acetic acid remains)
- mass CH3COONa produced = 0.05952 × 82.03 = 4.879 g
- mass H2O produced = 0.05952 × 18.02 = 1.071 g
- mass CO2 produced = 0.05952 × 44.01 = 2.618 g

Total mass of reactants consumed = 5.00 + 3.572 = 8.572 g = total mass of products formed (4.879 + 1.071 + 2.618 = 8.568 g, small rounding differences).

If CO2 escapes, mass left in container = sodium acetate + water + leftover acetic acid = 4.879 + 1.071 + 1.428 = 7.378 g.

If you give the specific starting masses (and whether your vinegar is pure acetic acid or e.g. 5% solution), I’ll compute the exact “missing mass” you need.