Problem 95 - Chemistry and Chemical Reactivity

Question

A Sodium bicarbonate and acetic acid react according to the equation \(\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \longrightarrow\) $$\mathrm{NaCH}_{3} \mathrm{CO}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$ What mass of sodium acetate can be obtained from mixing \(15.0 \mathrm{g}\) of \(\mathrm{NaHCO}_{3}\) with \(125 \mathrm{mL}\) of \(0.15 \mathrm{M}\) acetic acid?

Step-by-Step Solution

Answer

1.54 g of sodium acetate is produced.

Step 1: Determine Moles of Reactants

First, calculate the moles of sodium bicarbonate \( \mathrm{NaHCO}_3 \). The molar mass of \( \mathrm{NaHCO}_3 \) is approximately \( 84.01 \text{ g/mol} \). The moles of \( \mathrm{NaHCO}_3 \) can be found using the formula:\[\text{moles of } \mathrm{NaHCO}_3 = \frac{15.0 \text{ g}}{84.01 \text{ g/mol}} = 0.1785 \text{ mol}\]Next, calculate the moles of acetic acid \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \). Given the molarity and volume:\[\text{moles of } \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} = 0.15 \text{ M} \times 0.125 \text{ L} = 0.01875 \text{ mol}\]

Step 2: Identify the Limiting Reactant

To determine the limiting reactant, compare the mole ratio from the balanced equation. The balanced equation suggests a 1:1 mole ratio between \( \mathrm{NaHCO}_3 \) and \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \).Since there are only 0.01875 moles of \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \) and 0.1785 moles of \( \mathrm{NaHCO}_3 \), \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \) is the limiting reactant.

Step 3: Calculate Mass of Sodium Acetate Produced

With \( 0.01875 \text{ mol} \) of the limiting reactant \( \mathrm{CH}_3\mathrm{CO}_2\mathrm{H} \), the same moles of \( \mathrm{NaCH}_3\mathrm{CO}_2 \) is produced, as the reaction is 1:1.Find the molar mass of sodium acetate \( \mathrm{NaCH}_3\mathrm{CO}_2 \), which is approximately \( 82.03 \text{ g/mol} \):\[\text{mass of } \mathrm{NaCH}_3\mathrm{CO}_2 = 0.01875 \text{ mol} \times 82.03 \text{ g/mol} = 1.5381 \text{ g}\]

Step 4: Conclusion

The mass of sodium acetate \( \mathrm{NaCH}_3\mathrm{CO}_2 \) obtained is \( 1.5381 \text{ g} \). This result is contingent upon the mixture being allowed to fully react with the limiting reactant acetic acid being used up completely.

Key Concepts

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It serves as a vital tool to understand how substances interact within a reaction. In any chemical equation, you'll encounter the concept of a mole. This is a unit of measurement that helps in quantifying substances at a molecular level.

• Stoichiometry involves using balanced equations to determine the proportions of reactants and products.
• It relies on the conservation of mass, ensuring that the quantity of each element is the same before and after the reaction.
• The balanced equation acts as a recipe, indicating how much of each ingredient you need for the reaction to occur efficiently.

In this exercise, the equation provided explains exactly how sodium bicarbonate reacts with acetic acid to form sodium acetate, carbon dioxide, and water. The stoichiometry helps us to calculate exactly how much sodium acetate is formed from a given amount of reactants.

Limiting Reactant

The limiting reactant is the substance that gets used up first during a chemical reaction. Once it is completely consumed, the reaction stops. This concept is crucial because it determines the maximum amount of product that can be formed. Whether you consider baking cookies or conducting a chemical reaction, it's the same idea: the ingredient you have the least of will limit how much you can make.
In this reaction, we compared sodium bicarbonate with acetic acid based on their mole ratios. Even though there was plenty of sodium bicarbonate, the small amount of acetic acid was the limiting reactant. This dictated that only 0.01875 moles of sodium acetate could be produced. Hence, identifying the limiting reactant in any problem is essential to predict the outcome of the reaction accurately.

Molar Mass

Molar mass is a fundamental concept in chemistry, representing the mass of one mole of a substance. It allows you to convert between the mass of a substance and the number of moles. The periodic table provides the atomic masses of elements, which can be used to find the molar mass.

• To calculate molar mass, add up the atomic masses of all the atoms in a molecule or compound.
• It is essential for converting grams to moles and vice versa.
• Used frequently in stoichiometry to relate masses of different reactants or products.

In this exercise, the molar mass of sodium bicarbonate was calculated as approximately 84.01 g/mol, while sodium acetate was found to be 82.03 g/mol. Using these values, we converted the mass of the reactants into moles to predict the amount of product produced.

Reaction Equation

A reaction equation is a symbolic representation of a chemical reaction. It shows the reactants (substances before the reaction) and the products (substances formed by the reaction) and includes their quantities.

• Balancing the equation is crucial to reflecting the law of conservation of mass.
• Coefficients in a balanced equation indicate the ratio of moles for each substance involved.
• Correctly written equations help in accurately determining the limiting reactant and the theoretical yield.

In this scenario, the reaction between sodium bicarbonate and acetic acid is depicted, leading to the formation of sodium acetate. The balanced equation is crucial as it simplifies the calculation of how much sodium acetate can be obtained from a given amount of reactants.