Problem 108 - Chemistry The Central Science In Si Units

Question

A fertilizer railroad car carrying \(129,840 \mathrm{~L}\) of commercial aqueousammonia (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?

Step-by-Step Solution

Answer

The mass of citric acid required is 128,902.8 kg.

Step 1: Determine the Mass of Aqueous Ammonia

First, convert the volume of aqueous ammonia from liters to cubic centimeters since density is given in \(\mathrm{g/cm}^3\). Use the conversion: \(1 \mathrm{~L} = 1000 \mathrm{~cm}^3\). Therefore, \(129,840 \mathrm{~L} = 129,840,000 \mathrm{~cm}^3\). Next, calculate the mass of the aqueous ammonia by multiplying the volume by the density: \(129,840,000 \mathrm{~cm}^3 \times 0.88 \mathrm{~g/cm}^3 = 114,259,200 \mathrm{~g}\).

Step 2: Determine the Mass of Ammonia in the Solution

Since the solution is 30% ammonia by mass, calculate the mass of ammonia using the mass found in step 1. The mass of ammonia is 30% of the total mass of the solution: \(114,259,200 \mathrm{~g} \times 0.30 = 34,277,760 \mathrm{~g}\).

Step 3: Determine the Moles of Ammonia

To find the moles of ammonia, use the molar mass of ammonia (\(\mathrm{NH}_3\)), which is approximately \(17.03 \mathrm{~g/mol}\). The number of moles of ammonia is \(\frac{34,277,760 \mathrm{~g}}{17.03 \mathrm{~g/mol}} = 2,013,426.9 \mathrm{~mol}\).

Step 4: Determine the Moles of Citric Acid Required

Each molecule of citric acid can donate three protons. Therefore, the number of moles of citric acid required is \(\frac{2,013,426.9 \mathrm{~mol}}{3} = 671,142.3 \mathrm{~mol}\).

Step 5: Calculate the Mass of Citric Acid Required

The molar mass of citric acid is approximately \(192.12 \mathrm{~g/mol}\). Therefore, the mass of citric acid required is \(671,142.3 \mathrm{~mol} \times 192.12 \mathrm{~g/mol} = 128,902,756 \mathrm{~g}\). Convert this to kilograms: \(128,902,756 \mathrm{~g} \times \frac{1 \mathrm{~kg}}{1000 \mathrm{~g}} = 128,902.8 \mathrm{~kg}\).

Key Concepts

Molar Mass

The molar mass of a substance is the mass of one mole of its particles. It expresses the relationship between the mass of a substance and the amount (in moles) of that substance, bridging macroscopic measurements and molecular quantities. The molar mass is expressed in grams per mole (g/mol). In the context of the exercise, we calculated the molar mass of ammonia (NH₃) and citric acid (C₆H₈O₇). For ammonia, the molar mass is approximately 17.03 g/mol. This value is calculated by adding the atomic masses of nitrogen and hydrogen from the periodic table.
The molar mass allows us to convert between grams and moles, which is crucial for stoichiometry problems. It helped us determine how many moles of ammonia were needed in the reaction, which then guided us in finding the required amount of citric acid.

Density

Density is a measure that tells us how much mass is contained in a given volume. It is typically expressed in grams per cubic centimeter (g/cm³) or kilograms per liter (kg/L). In the problem, we are given the density of the aqueous ammonia solution as 0.88 g/cm³. Density is used to convert between volume and mass, allowing us to find out exactly how much ammonia is present in a solution by mass.
Knowing the density allowed us to multiply the volume of the solution by its density to find the total mass of the ammonia solution spilled. This calculation was the first step in determining how much ammonia we had to neutralize.

Acid-base Neutralization

Acid-base neutralization is a chemical reaction where an acid and a base react to form water and a salt. In the context of this exercise, citric acid (an acid with three acidic protons per molecule) is used to neutralize the spilled ammonia, which is a base. The concept is essential in determining how much citric acid is needed to completely neutralize a given amount of ammonia.

When the ammonia and citric acid react, they combine in a stoichiometric ratio where one mole of ammonia reacts with one proton from citric acid. Since citric acid can donate three acidic protons, each mole of citric acid can neutralize three moles of ammonia. This allowed us to determine that the moles of citric acid required are one-third the moles of ammonia.

Mass Percent Composition

Mass percent composition expresses the concentration of a component in a mixture as a percentage of the total mass of the mixture. It tells us what fraction of a mixture's mass is due to a particular component. The exercise specifies that the aqueous ammonia solution is 30% ammonia by mass.
Knowing this mass percent composition is essential for calculating the mass of pure ammonia in the solution. By multiplying the total mass of the aqueous solution by the mass percent (0.30), we found the mass of ammonia alone, thus helping us to further calculate the moles of ammonia for the neutralization reaction. This step is crucial in accurately assessing how much acid is needed to complete the neutralization.