Question

NaHCO_3(s) + CH₃COOH_(aq) -----------> NaCH₃COO_(aq) + CO_2(g) + H₂O_(l)

You will be calculating the minimum mass of sodium hydrogen carbonate and the minimum volume of 3.0 M acetic acid required to completely fill a 1.0-L plastic bag at room conditions.

Assume the following: 100% reactant purity, complete reaction, and the barometric pressure and temperature are 772 mmHg and 20.5^oC.

(a) Calculate the number of moles of carbon dioxide gas that would fill the bag completely, using the ideal gas law.

(b) Calculate the minimum mass of sodium hydrogen carbonate needed to generate this amount of carbon dioxide.

(c) Calculate the minimum volume of 3.0 M acetic acid needed to generate this amount of carbon dioxide.

Answers (type just the numerical values with the proper amount of SFs):

a) _____ moles of carbon dioxide gas would fill the 1.0-L bag completely.

b)______ grams is the minimum mass of sodium hydrogen carbonate needed to generate the amount of carbon dioxide.

c) _______ mL is the minimum volume of 3.0 M acetic acid needed to generate the carbon dioxide.

Answer 1

a) 0.042 moles of carbon dioxide gas would fill the 1.0-L bag completely.

b) 3.5 grams is the minimum mass of sodium hydrogen carbonate needed to generate the amount of carbon dioxide.

c) 14 mL is the minimum volume of 3.0 M acetic acid needed to generate the carbon dioxide.

Explanation

According to ideal gas law,

moles CO₂ = [(pressure CO₂) * (volume CO₂)] / [(R) * (temperature CO₂)]

where

pressure CO₂ = barometric pressure

pressure CO₂ = 772 mmHg

pressure CO₂ = 772 mmHg * (1 atm / 760 mmHg)

pressure CO₂ = 1.0158 atm

volume CO₂ = volume of bag = 1.0 L

R = gas constant = 0.0821 L-atm/mol-K

temperature CO₂ = 20.5 ^oC = 293.5 K

Substituting the values,

moles CO₂ = [(1.0158 atm) * (1.0 L)] / [(0.0821 L-atm/mol-K) * (293.5 K)]

moles CO₂ = 0.042 mol

moles NaHCO₃ reacted = moles CO₂ formed

moles NaHCO₃ reacted = 0.042 mol

mass NaHCO₃ reacted = (moles NaHCO₃ reacted) * (molar mass NaHCO₃)

mass NaHCO₃ reacted = (0.042 mol) * (84.0 g/mol)

mass NaHCO₃ reacted = 3.54 g

moles CH₃COOH reacted = moles CO₂ formed

moles CH₃COOH reacted = 0.042 mol

volume CH₃COOH = (moles CH₃COOH reacted) / (concentration CH₃COOH)

volume CH₃COOH = (0.042 mol) / (3.0 M)

volume CH₃COOH = 0.014 L

volume CH₃COOH = 14.05 mL