Sodium hydrogen carbonate, NaHCO3 can be converted to sodium carbonate, Na2CO3, water, and carbon dioxide upon heating. a.) Provide the balanced equation for this reaction; you can use the delta sign to represent heat. b.) When 5.000 grams of NaHCO3 is heated, what is the theoretical yield of Na2CO3 that can be formed? c.) What is the percent yield if 3.00 grams of Na2CO3 was recovered after heating?
The correct answer and explanation is:
a. Balanced Chemical Equation:
2 NaHCO3→ΔNa2CO3+H2O+CO22 \, \text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2
b. Theoretical Yield of Na₂CO₃ from 5.000 g of NaHCO₃
- Molar Masses:
- NaHCO₃ = 84.01 g/mol
- Na₂CO₃ = 105.99 g/mol
- Convert mass of NaHCO₃ to moles:
5.000 g NaHCO384.01 g/mol=0.05952 mol NaHCO3\frac{5.000 \, \text{g NaHCO}_3}{84.01 \, \text{g/mol}} = 0.05952 \, \text{mol NaHCO}_3
- Use the stoichiometry (2 mol NaHCO₃ → 1 mol Na₂CO₃):
0.05952 mol NaHCO3×1 mol Na2CO32 mol NaHCO3=0.02976 mol Na2CO30.05952 \, \text{mol NaHCO}_3 \times \frac{1 \, \text{mol Na}_2\text{CO}_3}{2 \, \text{mol NaHCO}_3} = 0.02976 \, \text{mol Na}_2\text{CO}_3
- Convert moles of Na₂CO₃ to grams:
0.02976 mol×105.99 g/mol=3.155 g Na2CO30.02976 \, \text{mol} \times 105.99 \, \text{g/mol} = 3.155 \, \text{g Na}_2\text{CO}_3
Theoretical Yield = 3.155 g Na₂CO₃
c. Percent Yield Percent Yield=(Actual YieldTheoretical Yield)×100=(3.003.155)×100=95.1%\text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 = \left( \frac{3.00}{3.155} \right) \times 100 = 95.1\%
Explanation (300 words):
Sodium hydrogen carbonate (NaHCO₃), commonly known as baking soda, decomposes upon heating in a well-known thermal reaction. This decomposition produces sodium carbonate (Na₂CO₃), water (H₂O), and carbon dioxide (CO₂). The balanced equation, which follows the law of conservation of mass, shows that two moles of sodium bicarbonate yield one mole of sodium carbonate, one mole of water, and one mole of carbon dioxide gas.
To determine the theoretical yield of sodium carbonate, we start by converting the given mass of NaHCO₃ (5.000 grams) to moles using its molar mass (84.01 g/mol). The stoichiometric relationship from the balanced equation (2:1 ratio of NaHCO₃ to Na₂CO₃) allows us to determine the moles of Na₂CO₃ that could theoretically be produced. Converting those moles to grams using the molar mass of sodium carbonate (105.99 g/mol) gives a theoretical yield of 3.155 grams.
However, chemical reactions rarely proceed with 100% efficiency due to factors such as incomplete reactions, side reactions, or loss during product recovery. In this scenario, if only 3.00 grams of Na₂CO₃ are recovered after the heating process, we calculate the percent yield by dividing the actual yield by the theoretical yield and multiplying by 100. This gives a percent yield of approximately 95.1%, indicating a highly efficient reaction with minimal losses.
This analysis demonstrates fundamental stoichiometric principles and shows how theoretical and actual yields are used to evaluate the efficiency of chemical processes in both academic and industrial settings.