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 \(29.0 \mathrm{~g} / \mathrm{mol} .)\)

Short answer

To produce the maximum allowable concentration of \(\mathrm{H}_{2} \mathrm{S}(g)\) in a room measuring \(3.5 \mathrm{~m} \times 6.0 \mathrm{~m} \times 2.5 \mathrm{~m}\) at \(101.3 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\), 3.20 grams of FeS would be required to react with hydrochloric acid.

Step-by-step solution

Calculate the mass of air in the room

To find the mass of air in the room, we first need to find the volume of the room. We can find this by multiplying the length, width, and height given: Room volume = Length × Width × Height Room volume = 3.5 m × 6.0 m × 2.5 m Room volume = 52.5 m³ Now, let's convert the volume into liters by multiplying by 1000: Room volume = 52.5 m³ × 1000 L/m³ = 52500 L Next, we'll use the Ideal Gas Law PV = nRT to find the number of moles of air in the room. We are given the temperature (T) as 25°C, which needs to be converted to Kelvin: T = 25°C + 273.15 = 298.15 K R (ideal gas constant) = 0.08206 L atm/mol K, P (given pressure) = 101.3 kPa = 101.3/101.3 = 1 atm. Now, we can calculate the number of moles of air (n) in the room: PV = nRT n = PV / RT n = (1 atm × 52500 L) / (0.08206 L atm/mol K × 298.15 K) n = 2137.7 mol The molar mass of air is given as 29.0 g/mol, so we can find the mass of air in the room: Mass of air = (2137.7 mol) × (29.0 g/mol) = 61994.3 g

Calculate the mass of \(\mathrm{H}_{2} \mathrm{S}(g)\) allowed in the room

Now, let's find out how much \(\mathrm{H}_{2} \mathrm{S}(g)\) mass is allowed in the room, based on the given allowable concentration of 20 mg/kg of air: Mass of \(\mathrm{H}_{2} \mathrm{S}(g)\) allowed = 20 mg/kg × (61994.3 g / 1000 g/kg) Mass of \(\mathrm{H}_{2} \mathrm{S}(g)\) allowed = 1.2399 g

Calculate the moles of \(\mathrm{H}_{2} \mathrm{S}(g)\) produced

To find out how many moles of \(\mathrm{H}_{2} \mathrm{S}(g)\) are produced, we can use the molar mass of \(\mathrm{H}_{2} \mathrm{S}(g)\), which is 34.08 g/mol: Moles of \(\mathrm{H}_{2} \mathrm{S}(g)\) = Mass of \(\mathrm{H}_{2} \mathrm{S}(g)\) / Molar mass of \(\mathrm{H}_{2} \mathrm{S}(g)\) Moles of \(\mathrm{H}_{2} \mathrm{S}(g)\) = 1.2399 g / 34.08 g/mol = 0.0364 mol

Calculate the mass of FeS required

The balanced chemical reaction between FeS and HCl is: FeS + 2HCl → FeCl₂ + H₂S From the balanced equation, we can see that one mole of FeS produces one mole of \(\mathrm{H}_{2} \mathrm{S}(g)\). So, the amount of FeS required to produce 0.0364 mol of \(\mathrm{H}_{2} \mathrm{S}(g)\) is also 0.0364 mol. Now, we can find the mass of FeS required using its molar mass, which is 87.91 g/mol: Mass of FeS = Moles of FeS × Molar mass of FeS Mass of FeS = 0.0364 mol × 87.91 g/mol = 3.20 g Therefore, 3.20 grams of FeS would be required to react with hydrochloric acid to produce the maximum allowable concentration of \(\mathrm{H}_{2} \mathrm{S}(g)\) in the room.

Key concepts

Chemical Reactions

Chemical reactions are processes in which substances, known as reactants, are transformed into different substances, called products. Understanding the basic principles of chemical reactions is key to solving stoichiometry problems.
Consider the chemical equation from the exercise:

• FeS + 2HCl → FeCl₂ + H₂S

This equation is balanced, meaning the number of each type of atom is the same on both sides of the equation. This is important because the law of conservation of mass states that mass cannot be created or destroyed in a chemical reaction.
In stoichiometry, we use balanced chemical equations to determine the amount of reactants needed to produce a desired amount of product. Each coefficient in the equation represents the relative number of moles of each substance. For example, the equation tells us that one mole of FeS reacts with two moles of HCl to produce one mole of H₂S. This ratio helps us calculate the mass of a reactant required if we know how much product we need to produce.

Ideal Gas Law

The ideal gas law is a mathematical relationship that describes the behavior of an ideal gas. It is expressed by the equation:

• PV = nRT

Where:

• P is the pressure of the gas,
• V is the volume of the gas,
• n is the number of moles of gas,
• R is the ideal gas constant, and
• T is the temperature in Kelvin.

This law is useful when performing calculations related to gases under specific conditions such as pressure, volume, and temperature.
In our exercise, we used the ideal gas law to determine the number of moles of air within the room. We solved for n (moles) using the provided room volume, temperature, and atmospheric pressure. This allows us to connect physical conditions with chemical quantities, crucial for calculating gas concentrations like the allowable \(\mathrm{H}_{2} \mathrm{S}(g)\) levels.

Molar Mass Calculation

Calculating molar mass is an important step in stoichiometry. Molar mass gives the mass of one mole of a substance and is measured in g/mol. It serves as a bridge between the mass of a material and the number of moles, allowing for conversions between these two quantities.
The molar mass of a compound is calculated by adding up the atomic masses of the elements in a formula. For instance, in our exercise:

• The molar mass of H₂S is 34.08 g/mol.
• The molar mass of FeS is 87.91 g/mol.

To calculate the mass of a substance when given the number of moles, we multiply the number of moles by the molar mass. Conversely, to find the number of moles from a given mass, we divide the mass by the molar mass. For example, knowing the allowable mass of \( \mathrm{H}_{2} \mathrm{S}(g) \) in the room allows us to determine how much FeS is needed, as one mole of FeS yields one mole of \( \mathrm{H}_{2} \mathrm{S}(g) \). Understanding these concepts is key to solving problems that involve stoichiometric calculations.