Problem 76 - Chemistry

Question

The bad smell of water containing hydrogen sulfide can be removed by the action of chlorine. The reaction is $$\mathrm{H}_{2} \mathrm{~S}(a q)+\mathrm{Cl}_{2}(a q) \longrightarrow 2 \mathrm{HCl}(a q)+\mathrm{S}(s)$$ If the hydrogen sulfide content of contaminated water is 22 ppm by mass, calculate the amount of \(\mathrm{Cl}_{2}\) (in grams) required to remove all the \(\mathrm{H}_{2} \mathrm{~S}\) from \(2.0 \times 10^{2}\) gallons of water. ( 1 gallon \(=3.785\) L.)

Step-by-Step Solution

Answer

The amount of chlorine required to remove all the hydrogen sulfide from the water is approximately \((2.0 \times 10^{2} \times 3.785 \times 22 \times 10^{-3})/(34.08) \times 70.90\) grams.

Step 1: Determine the Mass of Hydrogen Sulfide

Firstly, we need to understand that the problem presents the concentration of \( \mathrm{H}_{2} \mathrm{~S}\) as 22 ppm by mass, which we can consider as 22 mg of \( \mathrm{H}_{2} \mathrm{~S}\) per 1 kg of water. Given that the volume of water is \(2.0 \times 10^{2}\) gallons, and knowing that 1 gallon equals \(3.785\) L, and that the density of water is approximated as 1 g/mL or 1 kg/L, we get that the total mass of water to be treated is \(2.0 \times 10^{2} \times 3.785\) kg. Therefore, the amount of \( \mathrm{H}_{2} \mathrm{~S}\) in this volume of water equals \(2.0 \times 10^{2} \times 3.785 \times 22 \times 10^{-3}\) g.

Step 2: Convert Mass to Moles

Next, we convert the mass of \( \mathrm{H}_{2} \mathrm{~S}\) to moles using its molar mass. The molar mass of \( \mathrm{H}_{2} \mathrm{~S}\) is \(1.01\) g/mol for hydrogen and \(32.06\) g/mol for sulfur, making a total of \(34.08\) g/mol. Therefore, the number of moles of \( \mathrm{H}_{2} \mathrm{~S}\) is \((2.0 \times 10^{2} \times 3.785 \times 22 \times 10^{-3})/(34.08)\) moles.

Step 3: Use Stoichiometry of Reaction

Given the balanced chemical equation, we have 1 mole of \( \mathrm{H}_{2} \mathrm{~S}\) reacting with 1 mole of \( \mathrm{Cl}_{2}\). So, the number of moles of \( \mathrm{Cl}_{2}\) required is the same as the number of moles of \( \mathrm{H}_{2} \mathrm{~S}\), which is \((2.0 \times 10^{2} \times 3.785 \times 22 \times 10^{-3})/(34.08)\) moles.

Step 4: Convert Moles to Mass

Finally, we convert the number of moles of \( \mathrm{Cl}_{2}\) to mass using its molar mass. The molar mass of \( \mathrm{Cl}_{2}\) is \(35.45\) g/mol for each chlorine atom, so for two atoms, it's \(70.90\) g/mol. Hence, the mass of \( \mathrm{Cl}_{2}\) required is the number of moles times the molar mass, \((2.0 \times 10^{2} \times 3.785 \times 22 \times 10^{-3})/(34.08) \times 70.90\) g.

Key Concepts

Chemical Equations

In chemistry, chemical equations represent a concise way to express chemical reactions. They show the reactants on one side and the products on the other, connected by an arrow indicating the direction of the reaction. In our chlorine and hydrogen sulfide reaction, the equation is straightforward: \( \text{H}_2 \text{S}(aq) + \text{Cl}_2(aq) \rightarrow 2 \text{HCl}(aq) + \text{S}(s) \). This indicates that when hydrogen sulfide reacts with chlorine, it produces hydrochloric acid and elemental sulfur. The balanced equation above helps maintain the law of conservation of mass, ensuring the same number of atoms for each element is present on both sides.

• Reactants: substances present before the reaction, in this case, hydrogen sulfide \(\text{H}_2 \text{S}\) and chlorine \(\text{Cl}_2\).
• Products: substances formed as a result of the reaction, here hydrochloric acid \(\text{HCl}\) and sulfur \(\text{S}\).

A balanced chemical equation is crucial for stoichiometry, which is the calculation of reactants and products in chemical reactions. It ensures each element's mass is accounted for in both reactants and products.

Concentration Calculations

Concentration calculations are essential in determining how much of a substance is present in a solution. In the given exercise, the concentration of hydrogen sulfide \(\text{H}_2 \text{S}\) is given as 22 ppm (parts per million). This can be understood as 22 milligrams of hydrogen sulfide per kilogram of water, which is a measure of concentration by mass.
Since concentration helps us understand how much solute is in a given amount of solvent, it forms the basis of further calculations to understand how much reagent is needed to achieve a desired reaction outcome.
In the exercise:

• The mass of water is calculated knowing water’s density is approximately 1 kg/L, coupled with the volume in gallons and converting to liters.
• By considering the ppm value, we multiply to find the total amount of \(\text{H}_2 \text{S}\) in grams.
• This conversion is done using the relation that 1 ppm equals 1 mg/liter or kg.

This step ensures we accurately determine the treatable amount of hydrogen sulfide in the entire body of water.

Molar Mass

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, measured in grams per mole (g/mol). It is crucial for transforming between mass and moles, an essential aspect of stoichiometry.
In our exercise, the molar masses involved are:

• \(\text{H}_2 \text{S}\): Calculated by adding the atomic weights of hydrogen (1.01 g/mol each) and sulfur (32.06 g/mol), making a total of 34.08 g/mol.
• \(\text{Cl}_2\): Since chlorine’s atomic weight is 35.45 g/mol, for a \(\text{Cl}_2\) molecule (two chlorine atoms), the molar mass is 70.90 g/mol.

Knowing these molar masses allows us to convert the mass of \(\text{H}_2 \text{S}\) into moles before proceeding with the stoichiometric calculations.
Generally, molar mass helps chemists determine how much of a chemical is needed or produced in a reaction by relating mass to the number of molecules (or moles) involved. It enables precise measurement and balancing of chemical equations, vital for any chemical synthesis or analysis.