How much sodium bicarbonate is needed to fully react with 25mL of vinegar?

The chemical reaction between sodium bicarbonate (NaHCO3) and vinegar (CH3COOH) is given as:

`NaHCO_3 + CH_3COOH -> CH_3COONa + H_2CO_3`

Thus, 1 mole of sodium bicarbonate reacts with 1 mole of vinegar.

The molar mass of sodium bicarbonate = 23 + 1 + 12 + 3 x 16 = 84 g/mol

The molar mass of vinegar = 2 x 12 + 4 x 1 + 2 x 16 = 60 g/mol

The density of vinegar is 1.05 g/cm^3 or 1.05 g/ml.

Here, we have 25 ml of vinegar or 25 x 1.05 g = 26.25 g of vinegar.

This is also equal to 26.25 g x (1 mol /60 g) = 0.4375 mol

From the balanced chemical equation, 1 mole of vinegar will react will 1 mole of sodium bicarbonate. 

Or, 0.4375 mol vinegar will react with 0.4375 mol sodium bicarbonate.

This is equivalent to 0.4375 mol x (84 g/mol) = 36.75 g.

Thus, 36.75 g of sodium bicarbonate is needed to react with 25 ml of vinegar.

Hope this helps.