Problem 108 - Chemistry The Central Science

Question

A fertilizer railroad car carrying \(34,300\) gallons of commercial aqueous ammonia \((30 \%\) ammonia by mass) tips over and spills. The density of the aqueous ammonia solution is 0.88 \(\mathrm{g} / \mathrm{cm}^{3} .\) What mass of citric acid, \(\mathrm{C}(\mathrm{OH})(\mathrm{COOH})\left(\mathrm{CH}_{2} \mathrm{COOH}\right)_{2},\) (which contains three acidic protons) is required to neutralize the spill? 1 gallon \(=3.785 \mathrm{L} .\)

Step-by-Step Solution

Answer

To neutralize the spilled 34,300 gallons of 30% aqueous ammonia solution, 128,913.12 grams of citric acid are required.

Step 1: Calculate the mass of spilled ammonia

First, let's convert the volume of the spilled aqueous ammonia solution into mass. We are given the density of the solution and can use the conversion factor 1 gallon = 3.785 L. The volume of the spill is 34,300 gallons, so: \(V_{solution} = 34,300 \,gallons * \frac{3.785 \,L}{1 \,gallon} = 129,854 \,L \) Now, we can convert the volume of the spilled solution into mass using the density: \(m_{solution} = V_{solution} * \rho \) \(m_{solution} = 129,854 \,L * 0.88 \frac{g}{cm^{3}} * \frac{1000 \,cm^3}{1 \,L} \) \(m_{solution} = 114,271.52 \,g \) Since the ammonia concentration is 30% by mass, the mass of the spilled ammonia is: \(m_{NH_{3}} = 0.30 * m_{solution} \) \(m_{NH_{3}} = 0.30 * 114,271.52 \,g = 34,281.456 \,g \)

Step 2: Calculate the moles of spilled ammonia

Next, let's convert the mass of spilled ammonia into moles. We know that the molar mass of ammonia is: M(NH3) = 14.01 g/mol (N) + 3 * 1.01 g/mol (H) = 17.03 g/mol Now, we can calculate the moles of ammonia: \(n_{NH_{3}} = \frac{m_{NH_{3}}}{M(NH_{3})} \) \(n_{NH_{3}} = \frac{34,281.456 \,g}{17.03 \frac{g}{mol}} \) \(n_{NH_{3}} = 2012.35 \,mol \)

Step 3: Calculate the moles of citric acid needed to neutralize the spill

Citric acid has three acidic protons, which means that one mole of citric acid neutralizes three moles of ammonia. Therefore, we can calculate the moles of citric acid needed: \(n_{citric\,acid} = \frac{n_{NH_{3}}}{3} \) \(n_{citric\,acid} = \frac{2012.35 \,mol}{3} \) \(n_{citric\,acid} = 670.78 \,mol \)

Step 4: Calculate the mass of citric acid needed

Finally, let's find the mass of citric acid required to neutralize the spilled ammonia. The molar mass of citric acid is: M(citric acid) = 12.01 g/mol (C) * 6 + 1.01 g/mol (H) * 8 + 16.00 g/mol (O) * 7 = 192.12 g/mol Now, we can calculate the mass of citric acid: \(m_{citric\,acid} = n_{citric\,acid} * M(citric\,acid) \) \(m_{citric\,acid} = 670.78 \,mol * 192.12 \frac{g}{mol}\) \(m_{citric\,acid} = 128,913.12 \,g \) So, 128,913.12 grams of citric acid are needed to neutralize the spilled aqueous ammonia solution.

Key Concepts

Stoichiometry

Stoichiometry is the mathematical relationship between the relative quantities of substances involved in chemical reactions. It is the heart of the chemical equations we encounter in chemistry. Understanding stoichiometry allows us to predict the amounts of products formed or reactants required in a chemical reaction.

Considering the problem at hand, stoichiometry plays a critical role in determining the amount of citric acid needed to neutralize the ammonia spill. The reaction requires one mole of citric acid to neutralize three moles of ammonia. Using the stoichiometric coefficients, we can set up a ratio that allows us to convert from moles of ammonia to moles of citric acid.

To improve the understanding of stoichiometry in exercises, visual aids such as reaction diagrams showing the mole-to-mole ratios can be beneficial. Additionally, highlighting the importance of balancing chemical equations to respect the law of conservation of mass can provide students with a clearer insight into stoichiometric calculations.

Molar Mass Calculation

Calculating molar mass is a fundamental concept in chemistry that involves determining the mass of one mole of a substance. The molar mass is the sum of the atomic masses of all atoms in a molecule, often measured in grams per mole (g/mol).

In our exercise, the calculation of molar mass is essential for both ammonia (NH₃) and citric acid (C₆H₈O₇). By finding the molar mass of ammonia, we establish the conversion factor between the mass of ammonia (in grams) and the quantity of ammonia (in moles). This conversion factor is then used to calculate the amount of citric acid needed to neutralize the ammonia spill.

Illustrating the method of adding up atomic masses from the periodic table to find molar masses can help make this concept easier to grasp. Ensuring that students understand that molar mass connects the microscopic level (atoms or molecules) to the macroscopic level (grams) will aid them in performing these types of calculations accurately.

Acid-Base Neutralization

Acid-base neutralization is a chemical reaction where an acid and a base react to form water and a salt. This type of reaction is crucial in various fields, from environmental science to industrial processes. Neutralization reactions are often employed to manage spills of hazardous materials, as they can mitigate the effects of corrosive substances.

The citric acid used in our example is a triprotic acid, meaning it can donate three hydrogen ions (H⁺) per molecule. Ammonia acts as a base and can accept a hydrogen ion. The stoichiometry of the neutralization reaction is based on the mole ratio established by the balanced chemical equation for the reaction. For educational purposes, elaborating on the concept of mole ratios and equivalence points in neutralization can be pivotal for students to comprehend how to calculate the amount of an acid or base required for neutralization.

Simplifying complex terms, such as 'triprotic', with definitions and related examples, and showcasing neutralization reactions in real-world scenarios, can engage students and deepen their understanding of these reactions and their importance.