NaHCO3 + HC2H3O2 --> NaC2H3O2 + H2O + CO2 If you have exactly 12.2g of baking soda, how much vinegar would you need to add to react all of the baking soda?

Question

NaHCO3 + HC2H3O2 → NaC2H3O2 + H2O +_CO2

 

If you have exactly 12.2g of baking soda, how much vinegar would you need to add to react all of the baking soda?

Answer 1

NaHCO₃ + HC₂H₃O₂ --> NaC₂H₃O₂ + H₂O + CO₂ ... balanced equation

molar mass NaHCO₃ = 84.0 g / mole

molar mass HC₂H₃O₂ = 60.1 g / mole

12.2 g NaHCO3 x 1 mol / 84.0 g x 1 mol HC2H3O2 / mol NaHCO3 = 0.145 moles HC₂H₃O₂ needed

This is the moles of acetic acid (HC₂H₃O₂) needed, but vinegar is a dilute solution of acetic acid. Since we don't know the precise %, nor do we know the density of the vinegar, we really can't go any further in the calculations.

You can convert moles HC₂H₃O₂ to grams by multiplying by 60.1 g / mol = 8.73 grams HC₂H₃O₂ , but again, this is grams of acetic acid, not grams of vinegar.