Chapter 22: Problem 76

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.)

Short Answer

Expert verified

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 by step solution

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.

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.

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.

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.

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Short Answer

Step by step solution

Determine the Mass of Hydrogen Sulfide

Convert Mass to Moles

Use Stoichiometry of Reaction

Convert Moles to Mass

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