NaHCO3 + HCl -> H2O + NaCl + CO2 Data: 59.07 g = Mass of Empty Beaker ; Mass of Beaker with NaHCO3 = 59.78 g ; Mass of Beaker the Following Day = 59.57 g Questions: 1. Use stoichiometry to calculate the theoretical mass of sodium chloride that should be obtained from the initial mass of sodium hydrogen carbonate used. 2. Calculate the mass of sodium chloride obtained in this experiment. 3. Determine the yield of the experiment. 4. Determine the percent error of the experiment. 5. Determine the theoretical amount of water and carbon dioxide that should be formed from the initial amount of sodium hydrogen carbonate. 6. Calculate the maximum amount of hydrochloric acid that should be used in order not to have an excess. 7. If the density of water is 1 g/mL and 78.0 mL of water was obtained in this experiment, how many grams of sodium hydrogen carbonate reacted?
The Correct Answer and Explanation is:
Given Reaction:
NaHCO₃ + HCl → H₂O + NaCl + CO₂
Provided Data:
- Mass of Empty Beaker = 59.07 g
- Mass of Beaker with NaHCO₃ = 59.78 g
- Mass of Beaker after Reaction (Following Day) = 59.57 g
1. Theoretical Mass of Sodium Chloride (NaCl) from Sodium Hydrogen Carbonate (NaHCO₃)
Molar Masses:
- NaHCO₃ = 84.01 g/mol
- NaCl = 58.44 g/mol
Mass of NaHCO₃ Used:
59.78 g – 59.07 g = 0.71 g
Moles of NaHCO₃:
0.71 g ÷ 84.01 g/mol ≈ 0.00845 mol
From the balanced equation, molar ratio is 1:1, so moles of NaCl = 0.00845 mol
Theoretical Mass of NaCl:
0.00845 mol × 58.44 g/mol ≈ 0.494 g
2. Mass of Sodium Chloride Obtained in the Experiment
Mass after Reaction = 59.57 g
Mass of Empty Beaker = 59.07 g
Mass of NaCl Obtained:
59.57 g – 59.07 g = 0.50 g
3. Yield of the Experiment
Actual Mass of NaCl = 0.50 g
Theoretical Mass of NaCl = 0.494 g
Percent Yield:
(0.50 g ÷ 0.494 g) × 100 ≈ 101.2%
Yield over 100% likely indicates measurement error or impurities
4. Percent Error of the Experiment
Percent Error:
|Measured Value – Theoretical Value| ÷ Theoretical Value × 100
|0.50 g – 0.494 g| ÷ 0.494 g × 100 ≈ 1.21%
5. Theoretical Amount of Water and Carbon Dioxide Formed
From Reaction Stoichiometry: 1 mol NaHCO₃ produces:
- 1 mol H₂O (18.02 g/mol)
- 1 mol CO₂ (44.01 g/mol)
Moles of NaHCO₃ Used: 0.00845 mol
Mass of H₂O Produced:
0.00845 mol × 18.02 g/mol ≈ 0.152 g
Mass of CO₂ Produced:
0.00845 mol × 44.01 g/mol ≈ 0.372 g
6. Maximum Amount of Hydrochloric Acid Required
Molar Mass of HCl = 36.46 g/mol
Reaction ratio is 1:1
Moles of HCl Required = 0.00845 mol
Mass of HCl Required:
0.00845 mol × 36.46 g/mol ≈ 0.308 g
7. Sodium Hydrogen Carbonate Reacted from Water Collected
Density of Water = 1 g/mL
Volume of Water = 78.0 mL
Mass of Water = 78.0 g
From Reaction: 1 mol NaHCO₃ produces 1 mol H₂O
Moles of H₂O:
78.0 g ÷ 18.02 g/mol ≈ 4.33 mol
Moles of NaHCO₃ Required: 4.33 mol
Mass of NaHCO₃ Required:
4.33 mol × 84.01 g/mol ≈ 363.83 g
Thus, approximately 363.83 g of sodium hydrogen carbonate reacted to produce 78.0 mL of water.
Explanation (Summary)
The experiment involved reacting sodium hydrogen carbonate with hydrochloric acid to form sodium chloride, water, and carbon dioxide. Using precise mass measurements and stoichiometry, we calculated the expected yields for each product. The observed mass of sodium chloride closely matched the theoretical value, suggesting the reaction proceeded efficiently, though minor errors or impurities may have contributed to a slightly higher observed mass. Additionally, from water production data, we inferred the quantity of sodium hydrogen carbonate involved, showcasing the power of stoichiometric relationships in practical laboratory scenarios.
