Question

When Acetic acid (CH₃COOH) reacts with sodium bicarbonate a neutralization reaction occurs: [20 pts]

(a) Write the balanced chemical equation for this reaction.

(b) If you are given 1.26 x 10² g of acetic acid and 1.68 x 10² g of sodium bicarbonate, what reagent would be limiting?

(c) What would be the theoretical yield of the product(s)?

(d) How many grams of excess reagent would be left over?

(e) If the reaction yielded 68.46 g of the gas, 27.52 g of the liquid, and 159.6 g of the salt, what would their percent yields be?

Answer 1

a) CH3COOH + NaHCO3 ----> CH3COO- + H2O + CO2 + Na+

b) Mass of CH3COOH = 126 g

Mass of NaHCO3 = 168 g

Molar Mass of CH3COOH = 60.05 g / mol

Molar Mass of NaHCO3 = 84 g / mol

=> Moles of CH3COOH = 126 / 60.05 = 2.1

Moles of NaHCO3 = 168 / 84 = 2

CH3COOH + NaHCO3 ----> CH3COO- + H2O + CO2 + Na+

According to the stoichiometry of the reaction, 1 mole of NaHCO3 reacts with 1 mole of CH3COOH.

=> 2 mole of NaHCO3 reacts with 2 mole of CH3COOH. After this reaction NaHCO3 gets completrly exhausted and some of CH3COOH still remains.

Therefore, sodium bicarbonate is the limiting reagent

c)

According to the stoichiometry of the reaction 2 moles of NaHCO3 would produce 2 moles of each of the following

(CH3COONa , H2O , CO2)

Mass of CH3COONa = 2 x 82 = 164 g

Mass of H2O = 2 x 18 = 36 g

Mass of CO2 = 2 x 44 = 88 g

d)

Moles of excess reagent (CH3COOH) left = 2.1 - 2 = 0.1 moles

=> Mass = 0.1 x 60.05 = 6 g

e) Gas is CO2, Liquid is H2O and salt is CH3COONa

% yield of gas = 68.46 x 100 / 88 = 77.8 %

% yield of liquid = 27.52 x 100 / 36 = 76.44 %

% yield of salt = 159.6 x 100 / 164 = 97.32 %