Question

Although elemental chlorine, \(\mathrm{Cl}_{2}\), is added to drinking water supplies primarily to kill microorganisms, another beneficial reaction that also takes place removes sulfides (which would impart unpleasant odors or tastes to the water). For example, the noxious-smelling gas hydrogen sulfide (its odor resembles that of rotten eggs) is removed from water by chlorine by the following reaction: $$\mathrm{H}_{2} \mathrm{S}(a q)+\mathrm{Cl}_{2}(a q) \rightarrow \mathrm{HCl}(a q)+\mathrm{S}_{8}(s) \quad$$ (unbalanced) What mass of sulfur is removed from the water when 50\. L of water containing \(1.5 \times 10^{-5} \mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{S}\) per liter is treated with \(1.0 \mathrm{g}\) of \(\mathrm{Cl}_{2}(g) ?\)

Short answer

The mass of sulfur removed from the water when treated with 1.0 g of Cl2 is about \(7.06 \times 10^{-4}\, \mathrm{g}\).

Step-by-step solution

Balance the chemical equation

The unbalanced equation is: \(\mathrm{H}_{2} \mathrm{S}(a q)+\mathrm{Cl}_{2}(a q) \rightarrow \mathrm{HCl}(a q)+\mathrm{S}_{8}(s)\) To balance the equation, first balance sulfur atoms: \(\underline{8} \mathrm{H}_{2} \mathrm{S}(a q)+\mathrm{Cl}_{2}(a q) \rightarrow \mathrm{HCl}(a q)+\mathrm{S}_{8}(s)\) Second, balance hydrogen atoms and chloride atoms (since the hydrogen and chloride are only in the form of HCl, ensure the same number of molecules): \(8 \mathrm{H}_{2} \mathrm{S}(a q)+\underline{16} \mathrm{Cl}_{2}(a q) \rightarrow \underline{16} \mathrm{HCl}(a q)+\mathrm{S}_{8}(s)\) The balanced equation is: \[8 \mathrm{H}_{2}\mathrm{S}(aq) + 16 \mathrm{Cl}_2(aq) \rightarrow 16 \mathrm{HCl}(aq) + \mathrm{S}_{8}(s)\]

Convert the mass of hydrogen sulfide in water into moles

Given 50 L of water containing \(1.5 \times 10^{-5} \mathrm{g}\) of \(\mathrm{H}_{2}\mathrm{S}\), we calculate the total mass of H2S in the water: Total mass of H2S = \(50 \cdot 1.5 \times 10^{-5} \: \mathrm{g}\) = \(7.5 \times 10^{-4} \: \mathrm{g}\) Now, we need to convert this mass into moles. The molar mass of \(\mathrm{H}_{2}\mathrm{S}\) is approximately 34.1 g/mol. Moles of H2S = total mass/molar mass = \(\frac{7.5 \times 10^{-4} \: \mathrm{g}}{34.1 \: \mathrm{g/mol}}\) Moles of H2S = \(2.2 \times 10^{-5}\, \mathrm{mol}\)

Convert the mass of chlorine gas into moles

Given mass of Cl2 = 1.0 g. The molar mass of Cl2 is approximately 70.9 g/mol. To find the moles of Cl2, use the formula: Moles of Cl2 = mass/molar mass = \(\frac{1.0 \: \mathrm{g}}{70.9 \: \mathrm{g/mol}}\) Moles of Cl2 = \(1.41 \times 10^{-2} \, \mathrm{mol}\)

Determine the limiting reactant and calculate moles of sulfur produced

To find the limiting reactant, compare the mole ratios of H2S and Cl2 with respect to the balanced equation: Mole ratio of H2S:Cl2 = \(\frac{2.2 \times 10^{-5}\, \mathrm{mol}}{1.41 \times 10^{-2} \, \mathrm{mol}}\) = \(1.56 \times 10^{-3}\) From the balanced equation, the mole ratio of H2S:Cl2 for the reaction is 8:16 or 1:2. Since the actual mole ratio of H2S:Cl2 in the water is less than the required mole ratio, H2S is the limiting reactant. Now, based on the balanced equation, 8 moles of H2S produce 1 mole of sulfur (S8). Moles of S8 produced = \( \frac{2.2 \times 10^{-5}\, \mathrm{mol}(\mathrm{H}_{2}\mathrm{S})}{8}\) Moles of S8 produced = \(2.75 \times 10^{-6}\, \mathrm{mol}\, \mathrm{S}_{8}\)

Convert the moles of sulfur produced into mass

Now that we have the moles of sulfur produced, convert this to mass. The molar mass of S8 is approximately 256.5 g/mol. Mass of sulfur produced = moles of S8 * molar mass of S8 = \(2.75 \times 10^{-6}\, \mathrm{mol} \cdot 256.5 \: \mathrm{g/mol}\) Mass of sulfur produced = \(7.06 \times 10^{-4}\, \mathrm{g}\) Therefore, the mass of sulfur removed from the water when treated with 1.0 g of Cl2 is about \(7.06 \times 10^{-4}\, \mathrm{g}\).

Key concepts

Chemical Reactions

Chemical reactions involve the rearrangement of atoms in substances to form new substances. In the given exercise, the chemical reaction happens when hydrogen sulfide (\(\text{H}_2\text{S}\)) reacts with chlorine (\(\text{Cl}_2\)) to produce hydrochloric acid (\(\text{HCl}\)) and sulfur (\(\text{S}_8\)). This is a specific type of chemical reaction meant to purify drinking water.
In this context, the purpose is to eliminate the foul-smelling hydrogen sulfide by converting it into more benign substances.
This is common in water treatment processes where chlorine is used not only as a disinfectant but also as a reactant to remove unwanted elements that could affect taste or odor.
Understanding the nature of chemical reactions helps in predicting the products and balancing the equations needed to describe them accurately.

Limiting Reactant

The concept of the limiting reactant helps determine how much product can be formed in a chemical reaction. It is the substance that gets completely used up first, stopping the reaction from continuing.
In the reaction between \(\text{H}_2\text{S}\) and \(\text{Cl}_2\), we have to compare the relative amounts of each reactant to determine which one limits the formation of products. From our calculations:

• The mole ratio as described by the balanced chemical equation is 8 moles of \(\text{H}_2\text{S}\) to 16 moles of \(\text{Cl}_2\), which simplifies to a 1:2 ratio.
• Since in this problem, \(\text{H}_2\text{S}\) has a smaller mole ratio compared to \(\text{Cl}_2\) available, it becomes the limiting reactant.

Understanding which reactant is limiting allows us to calculate the maximum amount of product that can be formed, ensuring efficient use of resources in chemical manufacturing processes.

Mass-to-Moles Conversion

Converting mass into moles is a fundamental step in stoichiometry. To perform this, we use the formula: \( \text{moles} = \frac{\text{mass in grams}}{\text{molar mass}} \).

• In the given exercise, converting 50 L of water with a concentration of \(1.5 \times 10^{-5} \text{ g of } \text{H}_2\text{S per liter}\) to moles involves determining the total mass and dividing by \(\text{H}_2\text{S}\)'s molar mass, 34.1 g/mol.
• Similarly, converting 1 g of \(\text{Cl}_2\) to moles uses its molar mass, 70.9 g/mol.

This conversion is crucial as it allows chemists to work in terms of molecules and atoms rather than bulk mass, providing a basis for predicting how much product will form.

Balancing Chemical Equations

Balancing chemical equations ensures that the same number of each type of atom is present on both sides of the equation, following the Law of Conservation of Mass. For the reaction in this exercise:

• We start with the initial unbalanced equation: \(\text{H}_2\text{S + Cl}_2 \rightarrow \text{HCl + S}_8\).
• The first step is balancing the sulfur atoms, which reveals the need for 8 molecules of \(\text{H}_2\text{S}\) and 1 molecule of \(\text{S}_8\) on each side.
• Next, by balancing the hydrogen atoms, we ensure that 16 molecules of \(\text{HCl}\) are formed.
• Finally, balancing the chlorine atoms requires 16 molecules of \(\text{Cl}_2\), fully balancing the reaction equation.

This step is vital as it allows us to establish correct stoichiometric ratios, enabling accurate calculations of reactants and products needed or produced in the reaction.