Experiment - Postlab Limiting Reagent

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Bellevue University *

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CHEM 115

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Chemistry

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Feb 20, 2024

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Name ______________________________ Date ___________ Experiment – Stoichiometry Determining the Limiting Reagent Using the Reaction of Sodium Bicarbonate with Acetic Acid Post-Lab Questions 1. Describe any significant differences in the four reactions due to the different proportions of reagents. The differences in the four reactions depends on the reaction conditions, the stoichiometry, reaction kinetics, and the presence of catalysts or inhibitors. 2. Write a balanced equation for the reaction of acetic acid with sodium hydrogen carbonate CH 3 COOH + NaH CO 3 →CH 3 COONNa + H 2 O + CO 2 3. Frequently when performing stoichiometric calculations you are given the known mass of one of the reagents. In this case, the reagent, acetic acid, is dissolved in water to make vinegar so it is more difficult to determine the actual mass of acetic acid. Therefore, for this calculation, it is given that the 60.0 mL of vinegar used for each reaction contains 0.049 mol of HC 2 H 3 O 2 . How many grams of sodium hydrogen carbonate will react with this amount of acid? Molar mass of NaHCO 3 = 22.99g/mol + 1.008g/mol + 12.01g/mol + 16.00g/mol + 3 × ( 16.00g/mol ) = 22.99g/mol + 1.008g/mol + 12.01g/mol + 16.00g/mol + 48.00g/mol = 84.02g/mol Mass of NaHCO 3 = Number of moles × Molar mass = 0.049mol × 84.02g/mol = 4.12g 4.12 grams of sodium hydrogen carbonate will react with 60.0 mL of Vinegar containing 0.049 moles of acetic acid. 4. Compare the number of grams of sodium hydrogen carbonate expected to react with 60 mL of vinegar (as calculated in question 3) with each of the reactions in your set of experiments. Use this comparison to explain your observations for each of the four reactions in terms of limiting reactant. In each case, state which reactant was the limiting reactant. Were any of the combinations at a stoichiometric equivalence? Reaction 1: HC 2 H 3 O 2 + NaHCO 3 → CH 3 COONa + H 2 O + CO 2 From the balanced equation, the stoichiometric ratio of HC 2 H 3 O 2 to NaHCO 3 is 1:1. So, the moles of NaHCO 3 required is also 0.049 moles. Let's denote the mass of NaHCO 3 required as m 1 Reaction 2: 2 HC 2 H 3 O 2 + Na 2 CO 3 → 2 CH 3 COONa + H 2 O + CO 2 From the balanced equation, the stoichiometric ratio of HC 2 H 3 O 2 to Na 2 CO 3 is 2:1. So, the moles of Na 2 CO 3 required is twice the moles of HC 2 H 3 O 2, which is 2 × 0.049 moles. Let's denote the mass of Na 2 CO 3 required as m 2 1

Name ______________________________ Date ___________ Reaction 3: 2 HC 2 H 3 O 2 + Na 2 SO 3 → 2 CH 3 COONa + H 2 O + SO 2 From the balanced equation, the stoichiometric ratio of HC 2 H 3 O 2 to Na 2 SO 3 is2:1 So, the moles of Na 2 SO 3 required is twice the moles of HC 2 H 3 O 2, which is 2 × 0.049 moles. Let's denote the mass of Na 2 SO 3 required as m 3 . Reaction 4 HC 2 H 3 O 2 + Na 2 SO 4 → CH 3 COONa + NaHSO 4 From the balanced equation, the stoichiometric ratio of HC 2 H 3 O 2 to Na 2 SO 4 is 1:1. So, the moles of Na 2 SO 4 required is also 0.049 moles. Let’s denote the mass of Na 2 SO 4 required as m 4 In reactions 2 and 3, NaHCO 3 is the limiting reactant. In reactions 1 and 4, the reactants are at stoichiometric equivalence. 2

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