Question

Calculate the number of moles of CO_2 that would be produced if 0.25 g of baking soda completely react with a stoichiometrically equivalent amount of acetic acid. Vinegar is a 5% (w/v) solution of acetic acid (CH_3CO_2H) in water. How many moles of acetic add are contained in 1.00 mL of vinegar? Calculate the number of moles of CO_2 that would be produced if 1.00 mL of vinegar completely reacts with a stoichiometrically equivalent amount of baking soda.

Answer 1

1) given: 0.25 g of baking soda (NaHCO₃)

Molar mass of baking soda = 84 g/mol

No. of moles of baking soda = 0.25 g / 84 g/mol = 2.97 * 10^-3 mol

Reaction fo baking soda with acetic acid

NaHCO₃ + CH₃COOH ----> NaC₂H₃O₂  + H₂O + CO₂

1 mole of NaHCO₃ will react with stoichometric amount of acetic acid to produce 1 mole of CO₂

2.97 * 10^-3 mol of NaHCO₃ will produce  2.97 * 10^-3 mol of CO₂

Answer  2.97 * 10^-3 mol of CO₂

2) given vinegar is 5 % (w/v) it means 5 g of vinegar in 100 ml of solution

hence there will be 0.05 g in 1 ml of solution

molar mass of vinegar = 60 g/mol

so moles of Vinegar = 0.05 g /60 g/mol = 8.33 * 10^-4 mol of vinegar

Answer; 8.33 * 10^-4 mol of vinegar

3) accoriding to reaction stoichometry 1 mole of vinegar will produce 1 mole of CO₂

so 8.33 * 10^-4 mol of vinegar of will produce 8.33 * 10^-4 mol of CO₂

Answer :   8.33 * 10^-4 mol of CO₂