Adam Smith (pinecone@sowega.net)
- Pick out the target. You want cc CO2.
- List the given information. You have 1 g of NaHCO3. You have an excess of 5% acetic acid solution (which means every 100 g vinegar contains 5 g of acetic acid).
- Connect the given information with the target.
You're trying to convert g NaHCO3 into cc CO2. Whenever the problem involves a
connection between two different substances, you must have a mole-to-mole
relationship between the two to solve the problem. Write and balance an equation for the reaction
between acetic acid and sodium bicarbonate to get this relationship. Then
...so you've cut a tough problem into two easier ones: the g-to-mol conversion of NaHCO3, and the mol-to-cc conversion of CO2. The first conversion will require a molecular weight, and the second will involve a molar volume.1 g NaHCO3 
mol NaHCO3 mole-to-mole ratio
from balanced equationmol CO2 cc CO2 You can estimate the molar volume of the CO2 using the ideal gas law. But since the problem doesn't say anything about temperature and pressure, you'll have assume specific values for these; P = 1 atm and T = 298 K are reasonable conditions. Then the number of cubic centimeters per mole is
V/n = RT/P = (82.06 cc atm mol-1 K-1) × (298 K) / (1 atm) = 2.445388× 104 cc/mol. - Do the math. Set up a series of conversion factors so that units cancel to ultimately give you cc CO2.
1 g NaHCO3 ( 1 mol NaHCO3
84.0 g NaHCO3) ( 2.45 × 104 cc CO2
1 mol CO2) = 291 cc CO2 - Check the answer. You know that a mole of gas will occupy about 22.4 L at STP (and you're not far off from STP). Every mole of bicarbonate produces a mole of carbon dioxide. So multiplying moles of bicarbonate by 22.4 L should give you a rough estimate in liters.
You can follow the same general procedure to answer the second question. This time, no gases are involved.