Gas Forming Reactions – Stoichiometric Ratios
Na2CO3 + CH3COOH and NaHCO3 + CH3COOH
Chemicals and Equipment Needed
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1 M Acetic Acid – K1
Na2CO3 anhydrous– G1
NaHCO3 – G1
2-500 mL Erlenmeyer flasks – P1
o narrow mouth
Stoppers for flasks – U3
2 balloons – V3/4
Powder funnel – P3
2 Weighboats – A3
Scotch tape or parafilm
Preparation
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Add 100 mL 1 M acetic acid to each flask. Stopper and label.
Partially blow up the balloons and write “0.1 moles Na2CO3” on one and “0.1 moles NaHCO3 on the
other. Weigh out 10.6 g Na2CO3 and use the powder funnel to pour the powder into the appropriate
balloon. Weigh out 8.4 g NaHCO3 and pour into the other balloon.
On delivery: stretch the necks of the balloons over the flask openings and secure with tape or parafilm.
Presentation
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Lift up each balloon and shake the powder into the flask. Compare the final size of the balloons.
Discussion
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The balloon with NaHCO3 is larger due to the difference in stoichiometric ratios:
Na2CO3 (s) + 2 CH3COOH (aq) → CO2 (g) + H2O (ℓ) + 2 NaCH3COO (aq)
NaHCO3 (s) + CH3COOH (aq) → CO2 (g) + H2O (ℓ) + NaCH3COO (aq)
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Even though there are equal moles of Na2CO3 and CH3COOH in the first balloon, 1:2 ratio means the
acetic acid is the limiting reactant, so the amount of CO2 produced is half that of the second reaction,
where the reactants are present in stoichiometric amounts.
Clean-Up
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All solutions can go down the drain with plenty of water
Revised Spring 2015
AMM