Problem 105 - Chemistry and Chemical Reactivity

Question

A Sodium bicarbonate and acetic acid react according to the equation \(\mathrm{NaHCO}_{3}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \rightarrow\) $$ \mathrm{NaCH}_{3} \mathrm{CO}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ 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

1.537 grams of sodium acetate can be obtained.

Step 1: Find moles of NaHCO3

First, calculate the number of moles of sodium bicarbonate (\(\mathrm{NaHCO}_{3}\)). Use the molar mass of \(\mathrm{NaHCO}_{3}\), which is approximately \(84.01\ g/mol.\)\[\text{Moles of } \mathrm{NaHCO}_{3} = \frac{15.0\ g}{84.01\ g/mol} \approx 0.1785\ moles\]

Step 2: Find moles of CH3CO2H

Next, calculate the moles of acetic acid (\(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}\)) available in the solution using its molarity and volume. The given molarity is \(0.15\ M\), and the volume in liters is \(0.125\ L.\)\[\text{Moles of } \mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H} = 0.15\ mol/L \times 0.125\ L = 0.01875\ moles\]

Step 3: Determine the limiting reactant

The reaction is balanced and shows a 1:1 mole ratio between \(\mathrm{NaHCO}_{3}\) and \(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}\). With \(0.1785\) moles of \(\mathrm{NaHCO}_{3}\) and \(0.01875\) moles of \(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}\), acetic acid is clearly the limiting reactant because there are fewer moles of it.

Step 4: Calculate moles of NaCH3CO2 produced

The limiting reactant (\(\mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H}\)) determines the amount of product formed. Therefore, the moles of sodium acetate (\(\mathrm{NaCH}_{3}\mathrm{CO}_{2}\)) produced is equal to the moles of the limiting reactant.\[\text{Moles of } \mathrm{NaCH}_{3}\mathrm{CO}_{2} = 0.01875\ moles\]

Step 5: Convert moles of NaCH3CO2 to grams

Finally, calculate the mass of sodium acetate (\(\mathrm{NaCH}_{3}\mathrm{CO}_{2}\)) using its molar mass, which is approximately \(82.03\ g/mol.\)\[\text{Mass of } \mathrm{NaCH}_{3}\mathrm{CO}_{2} = 0.01875\ moles \times 82.03\ g/mol \approx 1.537\ g\]

Key Concepts

Limiting Reactant

In any chemical reaction, the limiting reactant is the substance that gets completely consumed first and thus determines the maximum amount of product that can be formed. It restricts the extent of the reaction. To identify the limiting reactant, it is essential to compare the moles of each reactant available with the moles required according to the balanced equation.
In the given reaction between sodium bicarbonate (\( \mathrm{NaHCO}_{3} \)) and acetic acid (\( \mathrm{CH}_{3}\mathrm{CO}_{2}\mathrm{H} \)), the equation is balanced and reflects a 1:1 mole ratio. However, only 0.01875 moles of acetic acid are present compared to 0.1785 moles of sodium bicarbonate. This indicates that acetic acid is the limiting reactant. Because it is smaller in moles than sodium bicarbonate, it determines the amount of sodium acetate (\( \mathrm{NaCH}_{3}\mathrm{CO}_{2} \)) that can be produced.
Knowing the limiting reactant is essential for calculating the maximum yield of a chemical reaction. It helps chemists to ensure efficient resource use and predict the quantity of product that can be obtained in a reaction.

Molar Mass Calculation

To work through stoichiometry problems like the one given in the exercise, it is vital to understand how to calculate the molar mass of a compound. Molar mass is the mass of one mole of a substance and is expressed in grams per mole (\( g/mol \)). Calculating molar mass involves summing the atomic masses of all the elements in a compound, based on their abundance in one formula unit.
For instance, to calculate the molar mass of sodium bicarbonate (\( \mathrm{NaHCO}_{3} \)), you must sum the atomic masses of sodium (Na), hydrogen (H), carbon (C), and oxygen (O) as follows:

• Sodium (Na): 22.99 g/mol
• Hydrogen (H): 1.01 g/mol
• Carbon (C): 12.01 g/mol
• Oxygen (O): 16.00 g/mol (multiplied by 3 since there are three O atoms)

Adding these gives:\[ 22.99 + 1.01 + 12.01 + (3 \times 16.00) = 84.01 \ g/mol \]This calculation is critical as it allows conversion between the mass of a substance and the number of moles, which is a fundamental component in stoichiometric calculations.

Balanced Chemical Equation

A balanced chemical equation is a vital tool in chemistry as it provides information about the reactants and products in a chemical reaction. It also indicates the proportions in which substances react and are produced.
To achieve a balanced equation, ensure that the number of each type of atom is the same on both the reactant and product sides of the equation. This reflects the law of conservation of mass, which states that matter cannot be created or destroyed in a chemical reaction.
In the provided equation:\[\mathrm{NaHCO}_{3}(\mathrm{aq}) + \mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}) \rightarrow \mathrm{NaCH}_{3} \mathrm{CO}_{2}(\mathrm{aq}) + \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{H}_{2} \mathrm{O}(\ell) \]it is balanced as the number of each type of atom (Na, C, H, and O) is equal on both sides.
Remember, balancing equations ensures the correct proportions of reactants and products are calculated, which is crucial for determining the limiting reactant and the maximum product yield in any stoichiometry problem.