Problem 56 - Chemistry An Atoms First Approach

Question

Complete and balance each of the following reactions. a. the reaction between sulfur dioxide gas and oxygen gas b. the reaction between sulfur trioxide gas and water c. the reaction between concentrated sulfuric acid and su\(\operatorname{crose}\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\)

Step-by-Step Solution

Answer

The short version of the answer for the three reactions is as follows: a. \(SO_2 (g) + O_2 (g) \rightarrow 2 SO_3 (g)\) b. \(SO_3 (g) + H_2O (l) \rightarrow H_2SO_4 (l)\) c. \(2 H_2SO_4 (l) + C_{12}H_{22}O_{11} (s) \rightarrow 12 CO_2 (g) + 18 H_2O (l) + 2 H_2S (g)\)

Step 1: Write the unbalanced chemical equation

SO2 (g) + O2 (g) → SO3 (g)

Step 2: Determine the number of atoms for each element

On the left side, we have one S atom, three O atoms. On the right side, we have one S atom and three O atoms.

Step 3: Balance the atoms

In this case, the equation is already balanced because there are the same number of atoms of each element on both sides.

Step 4: Verify that the equation is balanced

The balanced equation is: SO2 (g) + O2 (g) → 2 SO3 (g) b. Reaction between sulfur trioxide gas and water

Step 1: Write the unbalanced chemical equation

SO3 (g) + H2O (l) → H2SO4 (l)

Step 2: Determine the number of atoms for each element

On the left side, we have one S atom, four O atoms, and two H atoms. On the right side, we have one S atom, four O atoms, and two H atoms.

Step 3: Balance the atoms

In this case, the equation is already balanced because there are the same number of atoms of each element on both sides.

Step 4: Verify that the equation is balanced

The balanced equation is: SO3 (g) + H2O (l) → H2SO4 (l) c. Reaction between concentrated sulfuric acid and sucrose

Step 1: Write the unbalanced chemical equation

H2SO4 (l) + C12H22O11 (s) → CO2 (g) + H2O (l) + H2S (g)

Step 2: Determine the number of atoms for each element

On the left side, we have one S atom, 15 O atoms, 14 H atoms, and 12 C atoms. On the right side, we have one S atom, an unknown number of O atoms, an unknown number of H atoms, and an unknown number of C atoms.

Step 3: Balance the atoms

First, we will balance the C atoms. Since there are twelve C atoms in the compound sucrose, we need twelve CO2 molecules in the balanced equation. H2SO4 (l) + C12H22O11 (s) → 12 CO2 (g) + H2O (l) + H2S (g) Next, we will balance the H atoms. There are 22 H atoms in the compound sucrose and 4 H atoms in the products, so we need 18 H2O molecules in the balanced equation. H2SO4 (l) + C12H22O11 (s) → 12 CO2 (g) + 18 H2O (l) + H2S (g) Finally, we will balance the O atoms. There are 11 O atoms in the compound sucrose and 28 O atoms in the products, so we need one more H2SO4 molecule in the balanced equation. 2 H2SO4 (l) + C12H22O11 (s) → 12 CO2 (g) + 18 H2O (l) + 2 H2S (g)

Step 4: Verify that the equation is balanced

The balanced equation is: 2 H2SO4 (l) + C12H22O11 (s) → 12 CO2 (g) + 18 H2O (l) + 2 H2S (g)

Key Concepts

Balancing Equations

Balancing chemical equations is an essential skill in chemistry that ensures the law of conservation of mass is satisfied. This law states that matter cannot be created or destroyed, so the number of atoms for each element in the reactants must equal the number of atoms in the products. The process of balancing equations involves several steps.
First, write down the unbalanced equation with all reactants and products. Next, count the number of atoms of each element on both sides. Adjust the coefficients (the numbers in front of molecules or compounds) to balance the atoms for each element. Coefficients change the number of molecules but not the structure of the molecules themselves.
Begin with elements that appear in only one reactant and one product, then move to more complex molecules. Always verify your equation to ensure it is balanced by recounting each atom on both sides. Balancing equations may sometimes involve fractional coefficients, which can be easily converted to whole numbers by multiplying all coefficients by the same factor.

Sulfur Compounds

Sulfur compounds are involved in a wide range of chemical reactions and have important industrial and biological roles. For example, sulfur dioxide (SO extsubscript{2}) is a gas released by volcanic activity and industrial processes, playing a major role in the formation of acid rain. When it reacts with oxygen (O extsubscript{2}), it forms sulfur trioxide (SO extsubscript{3}), which is further used to produce sulfuric acid.
Sulfur trioxide can react with water to form sulfuric acid (H extsubscript{2}SO extsubscript{4}), a strong acid critical in fertilizers, manufacturing, and chemical syntheses. The versatile chemistry of sulfur compounds involves changes in oxidation states, making these reactions rich study cases for redox chemistry.

• Sulfur dioxide: SO extsubscript{2} - used in disinfectants and preservatives.
• Sulfur trioxide: SO extsubscript{3} - precursor to sulfuric acid.
• Sulfuric acid: H extsubscript{2}SO extsubscript{4} - widely used in industry and laboratories.

Understanding the reactions of sulfur compounds is crucial for students learning chemical reactions and environmental impact.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the balanced chemical equation and allows chemists to predict how much product will form from a given amount of reactant or vice versa. Stoichiometry involves using mole ratios derived from the coefficients of the balanced equation to calculate these amounts.
To apply stoichiometry, follow these steps:

• Write and balance the chemical equation.
• Convert quantities of known substances into moles.
• Use the mole ratio to convert moles of a given substance to moles of the desired substance.
• Convert moles back to desired units (e.g., grams, liters).

For instance, in the reaction with sulfuric acid and sucrose, stoichiometry helps in figuring out how much product like carbon dioxide ( ext{CO} extsubscript{2}) is produced from a given amount of sucrose. Calculations based on stoichiometry assume conditions of complete reaction with no side products, providing insights into the efficiency and yield of chemical processes.