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Chapter 4: Problem 86

Elemental chlorine is commonly used to kill microorganisms in drinking water supplies as well as to remove sulfides. For example, noxious-smelling hydrogen sulfide gas is removed from water via the following unbalanced chemical equation: \(\mathrm{H}_{2} \mathrm{S}(\mathrm{aq})+\mathrm{Cl}_{2}(\mathrm{aq}) \longrightarrow \mathrm{HCl}(\mathrm{aq})+\mathrm{S}_{8}(\mathrm{s})\) (a) Write a balanced equation for this reaction. (b) Determine the mass in grams of elemental sulfur, \(S_{8},\) which is produced when 50.0 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 \(\mathrm{Cl}_{2}\) (c) Calculate the percent yield of the reaction if \(5.8 \times 10^{-4} \mathrm{g}\) of \(\mathrm{S}_{8}\) is generated.

Short Answer

Expert verified
(a) Balanced equation: \(8\mathrm{H}_2\mathrm{S} + 8\mathrm{Cl}_2 \rightarrow 16\mathrm{HCl} + \mathrm{S}_8\). (b) Produced mass of \(\mathrm{S}_8\): \(7.04 \times 10^{-4}\ g\). (c) Percent yield: 82.39\%.

Step by step solution

01

Balance the Chemical Equation

First, we need to balance the chemical equation given: \[ \text{Unbalanced equation: } \mathrm{H}_2\mathrm{S} + \mathrm{Cl}_2 \rightarrow \mathrm{HCl} + \mathrm{S}_8 \]1. To balance the sulfur, we notice there is \(\mathrm{S}_8\) on the product side, so we put an 8 in front of \(\mathrm{H}_2\mathrm{S}\): \[ 8(\mathrm{H}_2\mathrm{S}) + \mathrm{Cl}_2 \rightarrow \mathrm{HCl} + \mathrm{S}_8 \]2. To balance the hydrogen, we now have 16 hydrogens on the reactant side, so we put a 16 in front of \(\mathrm{HCl}\): \[ 8(\mathrm{H}_2\mathrm{S}) + \mathrm{Cl}_2 \rightarrow 16(\mathrm{HCl}) + \mathrm{S}_8 \]3. Finally, we need to balance the chlorine. We see now that there are 16 chlorines in \(16(\mathrm{HCl})\) so we put 8 in front of \(\mathrm{Cl}_2\):\[ 8(\mathrm{H}_2\mathrm{S}) + 8(\mathrm{Cl}_2) \rightarrow 16(\mathrm{HCl}) + \mathrm{S}_8 \]The balanced equation is:\[ 8\mathrm{H}_2\mathrm{S} + 8\mathrm{Cl}_2 \rightarrow 16\mathrm{HCl} + \mathrm{S}_8 \]
02

Calculate Mass of Elemental Sulfur

Now, let's calculate the mass of \(\mathrm{S}_8\) produced.1. **Determine moles of \(\mathrm{H}_2\mathrm{S}\)**: - Given that each liter of water contains \(1.5 \times 10^{-5} \text{g of } \mathrm{H}_2\mathrm{S}\), the total mass of \(\mathrm{H}_2\mathrm{S}\) in 50.0 L is: \[ 50.0 \, \text{L} \times 1.5 \times 10^{-5} \, \text{g/L} = 7.5 \times 10^{-4} \, \text{g} \] - The molecular weight of \(\mathrm{H}_2\mathrm{S}\) is approximately 34.08 g/mol. - Moles of \(\mathrm{H}_2\mathrm{S}\): \[ \frac{7.5 \times 10^{-4} \, \text{g}}{34.08 \, \text{g/mol}} = 2.20 \times 10^{-5} \, \text{mol} \]2. **Determine moles of \(\mathrm{Cl}_2\)**: - \(1.0 \text{g of } \mathrm{Cl}_2\) with a molar mass of about 70.90 g/mol gives: \[ \frac{1.0 \, \text{g}}{70.90 \, \text{g/mol}} = 0.0141 \, \text{mol} \]3. **Identify the limiting reactant**: - According to the balanced equation, the ratio between \(\mathrm{H}_2\mathrm{S}\) and \(\mathrm{S}_8\) is 8:1. - We can use \(\mathrm{H}_2\mathrm{S}\) to confirm: - Since \(2.20 \times 10^{-5}\) mol \(\mathrm{H}_2\mathrm{S}\) corresponds to \(\frac{2.20 \times 10^{-5}}{8} = 2.75 \times 10^{-6} \, \text{mol of } \mathrm{S}_8\). - \(\mathrm{Cl}_2\) is in excess.4. **Calculate mass of \(\mathrm{S}_8\)**: - Molar mass of \(\mathrm{S}_8\) is approximately 256.52 g/mol. - Thus, mass of \(\mathrm{S}_8\): \[ 2.75 \times 10^{-6} \, \text{mol} \times 256.52 \, \text{g/mol} = 7.04 \times 10^{-4} \, \text{g} \]
03

Calculate Percent Yield

To find the percent yield, use the formula:\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \%\]- **Actual yield** is given as \(5.8 \times 10^{-4} \, \text{g}\).- **Theoretical yield** from our calculations is \(7.04 \times 10^{-4} \, \text{g}\).Plug in the values:\[ \text{Percent Yield} = \left( \frac{5.8 \times 10^{-4} \, \text{g}}{7.04 \times 10^{-4} \, \text{g}} \right) \times 100 \%\]\[ \text{Percent Yield} = 82.39 \% \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balancing Chemical Equations
Balancing chemical equations is like solving a puzzle to ensure that both sides have the same number of each type of atom. It's crucial because it maintains the law of conservation of mass, which states that mass can neither be created nor destroyed in a chemical reaction. In our example reaction of \( \text{H}_2\text{S} + \text{Cl}_2 \rightarrow \text{HCl} + \text{S}_8 \), we start by balancing each element one at a time.
  • First, consider sulfur (S). The product side contains \( \text{S}_8 \), meaning we need 8 sulfur atoms on the reactant side. So, we place an 8 in front of \( \text{H}_2\text{S} \).
  • Next is hydrogen (H). We now have 16 hydrogen atoms on the reactant side (from \( 8{H}_2\text{S} \)), so we balance the hydrogen on the product side by placing a 16 in front of \( \text{HCl} \).
  • Lastly, chlorine (Cl). With 16 chlorine atoms now required on the product side (from \( 16\text{HCl} \)), we simply place an 8 in front of \( \text{Cl}_2 \) to balance it.
This step ensures that the number of each atom type remains constant throughout the reaction, making it fully balanced.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It uses the balanced chemical equation to calculate the amounts of reactants needed or products formed. In our example, stoichiometry helps us figure out how much \( \text{S}_8 \) is produced when \( \text{H}_2\text{S} \) and \( \text{Cl}_2 \) react.
  • To find the amount of \( \text{S}_8 \) produced, we first determine the amount of \( \text{H}_2\text{S} \) and \( \text{Cl}_2 \) available by converting grams to moles.
  • Using the balanced equation \( 8\text{H}_2\text{S} + 8\text{Cl}_2 \rightarrow 16\text{HCl} + \text{S}_8 \), the stoichiometric coefficients indicate the ratios of reactants to products.
  • Given amounts of \( \text{H}_2\text{S} \) and \( \text{Cl}_2 \), stoichiometry allows us to determine the limiting reactant and, therefore, find out how much \( \text{S}_8 \) can theoretically be produced.
Understanding stoichiometry ensures precise calculations in chemical reactions, saving time and resources.
Percent Yield
The percent yield is the efficiency of a chemical reaction and is calculated by comparing the actual yield to the theoretical yield. It gives insight into the reaction's success. For the given reaction, it's crucial to determine how efficiently \( \text{S}_8 \) is produced.
  • The actual yield is the amount actually produced (e.g., \( 5.8 \times 10^{-4} \text{ g} \)).
  • The theoretical yield is the amount expected based on stoichiometric calculations (e.g., \( 7.04 \times 10^{-4} \text{ g} \)).
  • To find the percent yield, you use the formula: \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \right) \times 100 \% \]
  • In our example, the percent yield is computed as: \[ \frac{5.8 \times 10^{-4}}{7.04 \times 10^{-4}} \times 100 \% = 82.39\%\]
Percent yield informs us about the effectiveness and possible improvements needed in the reaction process.
Limiting Reactant
The limiting reactant is the substance in a chemical reaction that runs out first, limiting the amount of products formed. Identifying the limiting reactant is essential for maximizing reaction efficiency.
  • First, we calculate the moles of each reactant. Using their given amounts ensures accurate identification of which reactant is limited.
  • In our reaction, the stoichiometric relationship between \( \text{H}_2\text{S} \) and \( \text{S}_8 \) is 8:1. Thus, comparing moles of \( \text{H}_2\text{S} \) with \( \text{Cl}_2 \) helps determine the limiting reactant.
  • In this case, \( \text{H}_2\text{S} \) is the limiting reactant given that it will be completely consumed before \( \text{Cl}_2 \).
  • This concept is crucial as it influences the maximum amount of product that can be produced and helps in optimizing chemical processes.
Understanding the limiting reactant helps enhance efficiency and reduce waste in chemical reactions.

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Most popular questions from this chapter

Answer true or false. (a) Stoichiometry is the study of mass relationships in chemical reactions. (b) To determine mass relationships in a chemical reaction, you first need to know the balanced chemical equation for the reaction. (c) To convert from grams to moles and vice versa, use Avogadro's number as a conversion factor. (d) To convert from grams to moles and vice versa, use molar mass as a conversion factor. (e) A limiting reagent is the reagent that is used up first. (f) Suppose a chemical reaction between A and B requires 1 mol of \(A\) and 2 mol of \(B\). If 1 mol of each is present, then \(\mathrm{B}\) is the limiting reagent. (g) Theoretical yield is the yield of product that should be obtained according to the balanced chemical equation. (h) Theoretical yield is the yield of product that should be obtained if all limiting reagent is converted to product. (i) Percent yield is the number of grams of product divided by the number of grams of the limiting reagent times 100 (j) To calculate percent yield, divide the mass of product formed by the theoretical yield and multiply by 100

Heats of reaction are frequently measured by monitoring the change in temperature of a water bath in which the reaction mixture is immersed. A water bath used for this purpose contains 2.000 L of water. In the course of the reaction, the temperature of the water rose \(4.85^{\circ} \mathrm{C}\) (a) How many calories were liberated by the reaction? (b) If \(2 \mathrm{kg}\) of a given reactant is consumed in the reaction, how many calories are liberated for each kilogram?

Ammonia is prepared industrially by the reaction of nitrogen and hydrogen according to the following equation: \(\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\) If \(29.7 \mathrm{kg}\) of \(\mathrm{N}_{2}\) is added to \(3.31 \mathrm{kg}\) of \(\mathrm{H}_{2}\) (a) Which reactant is the limiting reagent? (b) How many grams of the other reactant are left over? (c) How many grams of \(\mathrm{NH}_{3}\) are formed if the reaction goes to completion?

The heat of combustion of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is \(670 \mathrm{kcal} /\) mol. The heat of combustion of ethanol, \(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}\), is 327 kcal/mol. The heat liberated by oxidation of each compound is the same whether it is burned in air or metabolized in the body. On a kcal/g basis, metabolism of which compound liberates more heat?

What is the mass in grams of each number of molecules of formaldehyde, \(\mathrm{CH}_{2} \mathrm{O} ?\) (a) \(100 .\) molecules (b) \(3000 .\) molecules (c) \(5.0 \times 10^{6}\) molecules (d) \(2.0 \times 10^{24}\) molecules
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