Problem 64 - Chemistry Matter and Change

Question

Balance the reaction of hydrogen sulfide with atmospheric oxygen gas. $$\mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction of hydrogen sulfide and oxygen gas is: \(\mathrm{2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(s) + 2H_2O(g)}\).

Step 1: Write the unbalanced equation

The unbalanced equation is given as: \(\mathrm{H_2S(g) + O_2(g) \rightarrow SO_2(s) + H_2O(g)}\)

Step 2: Count the number of each type of atom on both sides

Count the number of hydrogen, sulfur, and oxygen atoms on both sides of the equation: Reactants: Hydrogen: 2 Sulfur: 1 Oxygen: 2 Products: Hydrogen: 2 Sulfur: 1 Oxygen: 3

Step 3: Determine which elements are not balanced

Comparing the number of atoms for each element on both sides of the equation, we can see that oxygen is not balanced.

Step 4: Balance the oxygen atoms

To balance the oxygen atoms, we need to find the least common multiple (LCM) of the number of oxygen atoms on both sides of the equation. In this case, the LCM is 6. To balance the oxygen atoms, multiply the coefficient of the reactant \(\mathrm{O_2}\) by 3, and the coefficients of the products \(\mathrm{SO_2}\) and \(\mathrm{H_2O}\) by 2: \( \mathrm{H_2S(g) + 3O_2(g) \rightarrow 2SO_2(s) + 2H_2O(g)} \)

Step 5: Check for balanced atoms

Now, let's make sure that all atoms are balanced in the equation. Count the atoms for each element on both sides: Reactants: Hydrogen: 2 Sulfur: 1 Oxygen: 6 Products: Hydrogen: 2 Sulfur: 2 Oxygen: 6 The only discrepancy is with the sulfur atoms. To balance them, add a coefficient of 2 to the reactant \(\mathrm{H_2S}\).

Step 6: Write the balanced equation

Now that all the atoms are balanced, write the balanced chemical equation: \(\mathrm{2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(s) + 2H_2O(g)}\)

Key Concepts

Chemical Reactions

Chemical reactions are processes in which substances, called reactants, change into new substances, called products. When a reaction occurs, bonds between atoms in the reactants are broken, and new bonds are formed to create the products. This change involves rearrangements of atoms and can result in changes in energy. Typically, reactions are symbolized using a chemical equation, which shows the reactants on the left side and the products on the right side with an arrow in between.

In the example of hydrogen sulfide reacting with oxygen, we begin with \(\mathrm{H_2S} \) and \(\mathrm{O_2} \) as reactants and end up with \(\mathrm{SO_2} \) and \(\mathrm{H_2O} \) as the products. Each molecule participates in the reaction through interactions that often reveal intriguing aspects of their chemical nature, such as oxidation state changes, which will be discussed further.

Chemical reactions are fundamental in understanding the nature and behavior of different substances and have broad applications from industrial manufacturing to biological processes in living organisms.

Stoichiometry

Stoichiometry is like the recipe of chemistry. It refers to the calculation of reactants and products in chemical reactions and ensures the law of conservation of mass, where matter is neither created nor destroyed. It's crucial for making sure that the right amount of each substance is used in reactions, much like making sure you follow the exact recipe when baking a cake to get perfect results.

The balanced chemical equation gives the stoichiometric ratios between the substances involved. In the balanced equation \(\mathrm{2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(s) + 2H_2O(g)}\), the coefficients (2, 3, 2, 2) in front of each molecule tell us the proportions of each substance required: 2 moles of hydrogen sulfide react with 3 moles of oxygen to produce 2 moles of sulfur dioxide and 2 moles of water. This ensures that the equation adheres to the conservation laws, making stoichiometry essential to predicting the amounts of products formed in reactions.

Oxidation-Reduction Reactions

Oxidation-reduction reactions, or redox reactions, involve the transfer of electrons between substances. One substance loses electrons (oxidation) while another gains electrons (reduction). These reactions are central to many chemical processes, including energy production and metabolism in organisms.

In the context of the reaction given, hydrogen sulfide (\(\mathrm{H_2S} \)) is oxidized to sulfur dioxide (\(\mathrm{SO_2} \)), and oxygen (\(\mathrm{O_2} \)) is reduced to water (\(\mathrm{H_2O} \)). During this process:

• Oxidation: Sulfur goes from an oxidation state of -2 in \(\mathrm{H_2S} \) to +4 in \(\mathrm{SO_2} \)
• Reduction: Oxygen goes from an oxidation state of 0 in \(\mathrm{O_2} \) to -2 in \(\mathrm{H_2O} \)

Recognizing these changes is fundamental for understanding the mechanisms and effects of redox processes in both natural and industrial contexts.

Chemical Equation Balancing Steps

Balancing a chemical equation involves ensuring that the same number of each type of atom appears on both sides of the equation. This step-by-step process helps to accurately reflect the conservation of mass and characteristics of the reaction.

The approach usually includes:

• Writing the unbalanced equation using correct chemical formulas of reactants and products.
• Counting the number of each type of atom in both reactants and products.
• Adjusting coefficients (the numbers in front of molecules) to get equal numbers of each type of atom on both sides.
• Checking your work to ensure all atoms balance and adjusting as necessary.

In the hydrogen sulfide and oxygen gas reaction, we start with the unbalanced equation: \(\mathrm{H_2S(g) + O_2(g) \rightarrow SO_2(s) + H_2O(g)}\). Through logical adjustments focusing on oxygen and finally sulfur, we reach the balanced version: \(\mathrm{2H_2S(g) + 3O_2(g) \rightarrow 2SO_2(s) + 2H_2O(g)}\). Balancing is essential in representing the relationship between reactants and products accurately.