Problem 99 - Chemistry The Central Science In Si Units

Question

The maximum allowable concentration of \(\mathrm{H}_{2} \mathrm{~S}(g)\) in air is \(20 \mathrm{mg}\) per kilogram of air ( 20 ppm by mass). How many grams of FeS would be required to react with hydrochloric acid to produce this concentration at \(101.3 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\) in an average room measuring \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m} ?\) (Under these conditions, the average molar mass of air is \(29.0 \mathrm{~g} / \mathrm{mol} .)\)

Step-by-Step Solution

Answer

Approximately 5.37 grams of FeS are required to achieve the allowable H2S concentration in the room.

Step 1: Calculate the Volume of the Room

First, calculate the volume of the room using the given dimensions. The formula for the volume of a rectangular prism is \( V = ext{length} \times ext{width} \times ext{height} \). Here, the room dimensions are 3.5 m, 6.0 m, and 2.5 m. Therefore, the volume of the room is \( 3.5 \times 6.0 \times 2.5 = 52.5 \text{ m}^3 \).

Step 2: Convert Room Volume to Mass of Air

The room volume needs to be converted into the mass of air using the conditions provided. Given the average molar mass of air is 29.0 g/mol, convert the volume in m\(^3\) to liters (since 1 m\(^3\) = 1000 L) and apply the ideal gas law under the conditions provided to find the mass of air.

Step 3: Calculate Air Mass Using Ideal Gas Law

Using the ideal gas law, \( PV = nRT \), where \( P = 101.3 \text{ kPa} = 1.013 \times 10^5 \text{ Pa}\), \( R = 8.314 \text{ J/mol K} \), and \( T = 25 + 273 = 298 \text{ K} \), solve for \( n \). Thus, \( n = \frac{1.013 \times 10^5 \times 52.5 \times 1000}{8.314 \times 298} \). The calculated molarity \( n \) is the amount in moles of air in the room.

Step 4: Convert Moles of Air to Grams

Find the mass of air by multiplying the mole count by the molar mass: \( m = n \times 29 \). Assume that \( n \) has been computed in the previous step to give you the mass \( m \) in grams.

Step 5: Connect Air Mass to H2S ppm

Identify how much \( \text{H}_2\text{S} \) can be present with the given concentration of 20 mg/kg. Calculate this in grams using the mass of air you've obtained. For example, if air's mass is \( m \), then \( 20 \times 10^{-3} \times \frac{m}{1000} \) gives the allowable grams of \( \text{H}_2\text{S} \).

Step 6: Determine Amount of FeS Required

The chemical reaction \( \text{FeS} + 2\text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2\text{S} \) shows a 1:1 molar relationship between \( \text{FeS} \) and \( \text{H}_2\text{S} \). Calculate the molar amount of \( \text{H}_2\text{S} \) from the mass using its molar mass (34.1 g/mol). Consequently, the same moles of \( \text{FeS} \) are needed.

Step 7: Calculate Final Mass of FeS

The molar mass of \( \text{FeS} \) is roughly 87.9 g/mol. Using the number of moles from the previous step, multiply by the molar mass of \( \text{FeS} \) to get the required mass in grams.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is an important tool in chemistry for understanding the behavior of gases in various conditions. This law is expressed as \( PV = nRT \), where \( P \) is the pressure of the gas, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin. In practical terms, this equation allows us to relate the physical properties of gases and solve for an unknown when the other properties are known. In the given problem, we utilize the Ideal Gas Law to compute the number of moles of air in the room. By doing this, we establish a connection between volume and gas properties under specific conditions of temperature and pressure. It's essential to convert temperatures to Kelvin and to use consistent units for pressure and volume, typically Pascal for pressure and liters or cubic meters for volume. With the Ideal Gas Law, the mystery of how gases behave at different temperatures, volumes, and pressures becomes clear and calculable.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It allows chemists to predict how much product will form from a given amount of reactants. In this exercise, stoichiometry helps us to know how much FeS is required to produce a certain concentration of \( \text{H}_2\text{S} \).We begin with the balanced chemical equation: \( \text{FeS} + 2\text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2\text{S} \). This equation reveals that one mole of FeS will react with two moles of HCl to produce one mole of \( \text{H}_2\text{S} \). The stoichiometric coefficients (the numbers in front of compounds in the balanced equation) inform us about the proportion of reactants and products involved.Utilizing stoichiometry, we calculate the moles of \( \text{H}_2\text{S} \) needed to reach the maximum allowable concentration in the room. The calculated moles of \( \text{H}_2\text{S} \) directly inform us of the moles of FeS required due to their 1:1 relationship in the balanced equation. Stoichiometry is the heart of predicting outcomes in chemical reactions and is a fundamental principle in chemistry.

Molar Mass Calculations

Molar mass connects the number of moles of a substance to its mass in grams. It is essential in converting between these two quantities. The molar mass is calculated as the sum of the atomic masses of all atoms present in a molecule.In our scenario, we need to calculate the molar mass of both \( \text{H}_2\text{S} \) and \( \text{FeS} \). The molar mass of \( \text{H}_2\text{S} \) is given as 34.1 g/mol, while that of \( \text{FeS} \) is approximately 87.9 g/mol. This calculation is crucial for translating the number of moles obtained from stoichiometric ratios into a tangible mass of a chemical.By multiplying the moles derived from stoichiometry by the molar mass, we obtain the mass of each compound needed or produced in a reaction. Molar mass calculations bridge the theoretical world of moles to the practical applications involving actual measurable masses, enabling precise preparation of chemical reactions in lab settings.