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 \(=3.785 \mathrm{~L}\).

Short answer

To neutralize the spill, 129,048,640 g of citric acid is required.

Step-by-step solution

Convert the volume of the spilled solution into mass

To get the mass of the spilled solution, we need to convert the volume of the solution from gallons to liters and then to cubic centimeters. After that, we can multiply the volume by the density of the solution: \( mass\_of\_solution = volume \times density \) First, convert gallons to liters: \( 34,300 \, gallons \times \frac{3.785 \, L}{1 \, gallon} = 129,922.5 \, L \) Now convert liters to cubic centimeters: \( 129,922.5 \, L \times \frac{1000 \, cm^3}{1 \, L} = 129,922,500 \, cm^3 \) Finally, calculate the mass of the solution: \( 129,922,500 \, cm^3 \times 0.88 \frac{g}{cm^3} = 114,332,600 \, g \) So, the mass of the spilled solution is 114,332,600 g.

Calculate the mass of ammonia in the solution

We are given that the container was carrying 30% ammonia by mass. So, we can calculate the mass of ammonia in the solution as follows: \( mass\_of\_ammonia = 0.3 \times mass\_of\_solution \) \( mass\_of\_ammonia = 0.3 \times 114,332,600 \, g = 34,299,780 \, g \) So, the mass of ammonia in the solution is 34,299,780 g.

Convert the mass of ammonia into moles

To do this, we will use the molar mass of ammonia, which is approximately 17 g/mol. \( moles\_of\_ammonia = \frac{mass\_of\_ammonia}{molar\_mass\_of\_ammonia} \) \( moles\_of\_ammonia = \frac{34,299,780 \, g}{17 \frac{g}{mol}} = 2,017,634 \, mol \) So, there are 2,017,634 moles of ammonia in the solution.

Determine the required moles of citric acid to neutralize the spill

Since citric acid has three acidic protons, three moles of ammonia can neutralize one mole of citric acid. Therefore, we need to divide the moles of ammonia by three to find the number of moles of citric acid required. \( moles\_of\_citric\_acid = \frac{moles\_of\_ammonia}{3} \) \( moles\_of\_citric\_acid = \frac{2,017,634 \, mol}{3} = 672,545 \, mol \) So, 672,545 moles of citric acid are needed to neutralize the spill.

Calculate the mass of citric acid needed to neutralize the spill

Using the molar mass of citric acid (approximately 192 g/mol), we can now find the mass of citric acid required to neutralize the spill. \( mass\_of\_citric\_acid = moles\_of\_citric\_acid \times molar\_mass\_of\_citric\_acid \) \( mass\_of\_citric\_acid = 672,545 \, mol \times 192 \frac{g}{mol} = 129,048,640 \, g \) Hence, 129,048,640 g of citric acid is required to neutralize the spill.

Key concepts

Chemical Spill Neutralization

When a chemical spill occurs, it is crucial to neutralize the substance to prevent environmental damage and hazards to human health. Neutralization is a chemical reaction where an acid and a base react to form water and a salt, effectively reducing the harmful effects of the spill.

For the given problem, the spilled chemical is ammonia, which behaves as a base. Citric acid, a weak organic acid containing three acidic protons, is chosen to neutralize the ammonia. The goal is to ensure that all the ammonia reacts with the citric acid to achieve a state where the solution is no longer basic. This process involves molar relationships and conversions that consider the properties of the involved compounds. Understanding this principle is essential in various real-world applications, such as environmental engineering, where the neutralization of spills is a common task.

Molarity and Concentration

When dealing with solutions, the concentration is a measure of how much solute is dissolved in a given quantity of solvent. Molarity is one specific way to express concentration, defined as the number of moles of solute per liter of solution (mol/L). In our textbook example, converting the volume of the spill from gallons to liters was an important step before determining the amount of citric acid needed for neutralization.

The solution's molarity could be deduced if desired, by dividing the number of moles of ammonia by the volume of the solution in liters. This information is pivotal when preparing solutions in a lab or scaling up the process for industrial applications. Accurate calculations ensure that reactions proceed in the desired way, which is particularly important in scenarios such as neutralizing a chemical spill.

Molar Mass

Understanding molar mass is vital in stoichiometry, which is the study of the quantitative relationships between the amounts of reactants and products in chemical reactions. Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). It acts as a conversion factor that helps chemists determine how much of a substance is needed to react with a given amount of another substance.

In our exercise, the molar mass of ammonia (17 g/mol) and citric acid (192 g/mol) allow us to convert between grams and moles to find exactly how much citric acid is required to neutralize the specified amount of ammonia. Without knowledge of molar masses, it would not be possible to carry out the precise calculations necessary for solving chemical problems, whether they occur in a controlled environment like a lab, or in emergency situations like the chemical spill described in the textbook.