Question

The maximum allowable concentration of \(\mathrm{H}_{2} \mathrm{~S}(\mathrm{~g})\) 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 \(1.00 \mathrm{~atm}\) and \(25^{\circ} \mathrm{C}\) in an average room measuring \(2.7 \times 4.3 \times 4.3 \mathrm{~m} ?\) (Under these conditions, the average molar mass of air is \(29.0 \mathrm{~g} / \mathrm{mol}\).)

Short answer

The mass of FeS required to react with hydrochloric acid to produce the maximum allowable concentration of H2S in the average room under the given conditions is 3.051 grams.

Step-by-step solution

Calculate mass of H2S to be produced

First, we must calculate the mass of air in the room to find out how many grams of H2S should be produced. The volume of the room is given by: 2.7 m × 4.3 m × 4.3 m, and we must convert the volume to liters, as 1 m^3 equals 1000 L. Volume = (2.7 m × 4.3 m × 4.3 m) × 1000 L/m³ = 49,797 L Now, we'll use the ideal gas law to find the mass of air in the room: PV = nRT => m = n * M Rearrange for moles (n): n = PV / RT Use the given pressure (P) of 1.00 atm, converted to L * kPa / (mol * K) units using the conversion factor 101.325 kPa/atm, the given temperature (T) of 25°C, which is 298 K when converted to kelvin, and the molar gas constant (R) as 8.314 L * kPa / (mol * K). n = (1.00 atm * 49,797 L * (101.325 kPa/atm)) / (8.314 L * kPa / (mol * K) * 298 K) = 2041.3 mol. Next, we will find the mass of the air in the room using the average molar mass of air (29.0 g/mol): m_air = n * M = 2041.3 mol * 29.0 g/mol = 59,197.7 g Now that we have the mass of the air in the room, we can determine the mass of H2S that is allowed based on the given ratio (20 mg/kg or 20 ppm by mass). We can convert g to mg and kg, then use cross-multiplication to find the required mass of H2S: m_H2S = (m_air * (20 mg/kg)) / (1 kg) = (59,197.7 g * 1000 mg/g * (20 mg/kg)) / (1000 g/kg) = 1183.95 mg

Use stoichiometry to find mass of FeS required

In the reaction between FeS and hydrochloric acid, H2S is produced. The balanced chemical equation is: FeS (s) + 2 HCl (aq) -> FeCl2 (aq) + H2S (g) From the balanced equation, we can see that 1 mole of FeS reacts to produce 1 mole of H2S. We must first convert the mass of H2S in mg to moles, using the molar mass of H2S (34.08 g/mol): n_H2S = (1183.95 mg * 1 g/1000 mg) / (34.08 g/mol) = 0.03472 mol Since the ratio of FeS to H2S is 1:1 in the balanced equation, the moles of FeS required is equal to the moles of H2S. n_FeS = n_H2S = 0.03472 mol

Convert the mass of FeS to grams

Now that we have the moles of FeS required, we can find the mass of FeS using the molar mass of FeS (87.91 g/mol): m_FeS = n_FeS * M_FeS = 0.03472 mol * 87.91 g/mol = 3.051 g So, 3.051 grams of FeS would be required to react with hydrochloric acid to produce the maximum allowable concentration of H2S in the average room under the given conditions.

Key concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that describes the behavior of ideal gases. It combines several physical laws into one equation: \[ PV = nRT \] 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.

This equation helps us understand how gases behave under different conditions.
In the exercise, we use the ideal gas law to calculate the number of moles of air in the given room by rearranging to solve for \(n\): \[ n = \frac{PV}{RT} \] We input the known values:

• Pressure \(P = 1.00\) atm, which is converted to kPa by multiplying by 101.325,
• Volume \(V = 49,797\) L (based on room dimensions), and
• Temperature \(T = 25^{\circ} C\), converted to 298 K.

Using these, we determine the moles of air to further calculate the mass, providing a foundational understanding of gas behavior in stoichiometric calculations.

Molar Mass

Molar mass is crucial for converting between the mass of a compound and the moles of that compound. It is the mass of one mole of a substance and is expressed in grams per mole (g/mol). Understanding molar mass is essential in stoichiometry because it allows conversion between mass and moles, making it easier to perform calculations involving chemical equations.
For example, in the problem, we calculate the molar mass of \(\text{H}_2\text{S}\) and \(\text{FeS}\) to convert between mass and moles. - The molar mass of \(\text{H}_2\text{S}\) is 34.08 g/mol. This value helps us convert the allowable mass of \(\text{H}_2\text{S}\) into moles to proceed with reaction calculations.- Similarly, the molar mass of \(\text{FeS}\) is 87.91 g/mol, allowing us to determine how many grams are needed once we know the number of moles required.Molar mass bridges the gap between actual weight and abstract moles, facilitating precise stoichiometric conversions in reaction scenarios.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products, often governed by a balanced chemical equation. In stoichiometry, understanding these reactions is key to determining the amount of each substance involved.
In this exercise, we observe the reaction between iron(II) sulfide (\(\text{FeS}\)) and hydrochloric acid (\(\text{HCl}\)), which produces hydrogen sulfide gas (\(\text{H}_2\text{S}\)) and iron(II) chloride (\(\text{FeCl}_2\)):\[ \text{FeS} + 2\text{HCl} \rightarrow \text{FeCl}_2 + \text{H}_2\text{S} \] This balanced equation shows us the ratio of reactants and products. Each mole of \(\text{FeS}\) produces one mole of \(\text{H}_2\text{S}\). By knowing this ratio, we deduce the exact amounts needed or produced during the reaction.Chemical reactions like this guide our steps in calculating the quantities of reactants and products, ensuring efficient and effective stoichiometry calculations based on balanced equations.

Concentration Calculations

Concentration calculations describe how much solute is present within a given amount of solvent or solution. It is often expressed in parts per million (ppm), which is crucial for determining allowable exposure levels in contexts such as environmental and industrial chemistry.
In the problem, the concentration of hydrogen sulfide (\(\text{H}_2\text{S}\)) gas is restricted to 20 ppm by mass. This translates to 20 mg of \(\text{H}_2\text{S}\) per 1 kg of air. To determine how much \(\text{H}_2\text{S}\) is permissible in our calculated mass of air (59,197.7 g), we use the ppm notation to find the maximum mass of \(\text{H}_2\text{S}\) allowed. This involves a straightforward conversion:

• \(\text{m}_{\text{H}_2\text{S}} = \text{mass of air} \times \frac{20 mg}{1 kg}\)

Being adept at these concentration calculations helps navigate thresholds and maintain safe, compliant chemical environments. It ensures precision and adherence to standards applicable in various applications such as laboratory safety and pollution assessment.