Problem 3 - Introductory Chemistry Atoms First

Question

\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \rightarrow \mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\)

Step-by-Step Solution

Answer

The balanced chemical equation is: \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\]

Step 1: Balance the Carbon atoms

There are 6 Carbon atoms on the left side and 1 on the right side. To balance the Carbon atoms, we need to put a coefficient of 6 in front of \(\mathrm{CO}_{2}\): \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+\mathrm{H}_{2} \mathrm{O}\] Now we have 6 carbon atoms on both sides of the equation.

Step 2: Balance the Hydrogen atoms

There are 12 Hydrogen atoms on the left side and 2 on the right side. To balance the Hydrogen atoms, we need to put a coefficient of 6 in front of \(\mathrm{H}_{2} \mathrm{O}\): \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\] Now we have 12 hydrogen atoms on both sides of the equation.

Step 3: Balance the Oxygen atoms

Finally, we need to balance the Oxygen atoms. There are 8 Oxygen atoms on the left side (6 in \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) and 2 in \(\mathrm{O}_{2}\)) and 18 on the right side (12 in 6\(\mathrm{CO}_{2}\) and 6 in 6\(\mathrm{H}_{2} \mathrm{O}\)). To balance the Oxygen atoms, we need to put a coefficient of 6 in front of \(\mathrm{O}_{2}\): \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\] Now we have 18 oxygen atoms on both sides of the equation.

Step 4: Verify that the chemical equation is balanced

To ensure the chemical equation is balanced, we can compare the number of atoms of each element in the reactants and the products: - Carbon: 6 on both sides - Hydrogen: 12 on both sides - Oxygen: 18 on both sides The chemical equation is balanced: \[\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\]

Key Concepts

Chemical Reactions

A chemical reaction involves the transformation of one or more substances into new substances. In the equation \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\), we observe the conversion of glucose \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) and oxygen \(\mathrm{O}_{2}\) into carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2} \mathrm{O}\). This type of reaction is known as a combustion reaction, in which a substance reacts with oxygen, releasing energy. In any chemical reaction, the reactants are the starting substances, and the products are the substances formed following the chemical change. Here, glucose and oxygen are reactants while carbon dioxide and water are products.It is crucial to represent a chemical reaction with a balanced equation. This shows the reactants and products as well as the proportions, ensuring that the law of conservation of mass is followed.

Stoichiometry

Stoichiometry is the part of chemistry that involves calculating the quantities of reactants and products in a chemical reaction. The coefficients in a balanced chemical equation like \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\) indicate the stoichiometric ratios. These ratios tell us how much of each substance is involved in the reaction. By looking at the coefficients:

• 6 moles of \(\mathrm{O}_{2}\) are required to react with 1 mole of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\).
• The reaction produces 6 moles of \(\mathrm{CO}_{2}\) and 6 moles of \(\mathrm{H}_{2} \mathrm{O}\).

Stoichiometry helps us predict how much product will be formed and how much reactant will be consumed. It is essential in various applications, including manufacturing, environmental science, and pharmaceuticals, to optimize the use of resources and energy.

Conservation of Mass

The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. This principle is fundamental for balancing chemical equations.In our balanced chemical equation \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6\mathrm{O}_{2} \rightarrow 6\mathrm{CO}_{2}+6\mathrm{H}_{2} \mathrm{O}\), we verify that the number of each type of atom is the same on both sides of the equation. This ensures that the mass of the reactants equals the mass of the products. After balancing:

• Carbon atoms: 6 on both sides
• Hydrogen atoms: 12 on both sides
• Oxygen atoms: 18 on both sides

Ensuring equal atoms on both sides reflects the conservation of mass, making the equation valid and showing the transformation complies with natural law. This principle is not only critical in chemistry but also applied broadly in physics and engineering to analyze systems' behavior.