Introduction: Stoichiometry is the calculation of the quantities of chemical elements or compounds involved in chemical reactions. It is the amount of substances used before the reaction and the product of the reaction. The Law of Conservation of Mass states that matter can be changed from one form into another, mixtures can be separated or made, and pure substances can be decomposed, but the total amount of mass remains constant. The ideal gas law is “a law relating the pressure, temperature, and volume of an ideal gas. Many common gases exhibit behavior very close to that of an ideal gas at ambient temperature and pressure. The ideal gas law was originally derived from the experimentally measured Charles' law and Boyle's law. Let P be the …show more content…
Pour 60 mL of distilled water into the beaker using a graduated cylinder. Carefully place 7.5 grams of acetic acid into the beaker. Place 10.5 grams of sodium bicarbonate into one of the fingers in the rubber glove. Use a rubber band to tie the finger shut. Fasten the rubber glove to the opening of the beaker using at least two rubber bands. When the rubber glove is securely fastened to the beaker,untie the finger of the glove and let the sodium bicarbonate fall into the beaker. Record your observations. When the reaction stops, find the volume of the rubber glove by submerging only the inflated glove into the overflowing bucket of water. Dispose of the glove. Refill the bucket and record how many liters it takes to refill it. That will equal the volume of the glove. Finally, wash the residue from the beaker, empty the bucket and clean any spilled solution from the …show more content…
Quantitative Results: The weight of the 250 mL beaker was 159.44 grams. 7.58 grams of acetic acid was used and 10.5 grams of sodium bicarbonate was used. The volume of the CO2 in the rubber glove was 4 liters. The pressure of the classroom was close to 1 atm. Calculations/Analysis We predicted that we should end up with .125 grams of carbon dioxide using the equation below. 7.5g C2H3O2H 1molC2H3O2H 1molCO2 .125 mol of CO2 60.0416g C2H3O2H 1molC2H3O2H Our percent yield was 133%. Our theoretical yield was calculated using the below equation. 101.3kPaL=(.125molCO2)*(8.31LkPaK mol)*(295K) L=3.025L of CO2.