Chapter 1: Problem 23
Count the atoms on both sides of the arrow to demonstrate that these equations are balanced. a. \(2 \mathrm{C}_{3} \mathrm{H}_{8}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}(g)+8 \mathrm{H}_{2} \mathrm{O}(l)\) b. \(2 \mathrm{C}_{8} \mathrm{H}_{18}(g)+25 \mathrm{O}_{2}(g) \longrightarrow 16 \mathrm{CO}_{2}(g)+18 \mathrm{H}_{2} \mathrm{O}(l)\)
Short Answer
Step by step solution
Understand the Equation
Count Atoms in Equation (a) - Left Side
Count Atoms in Equation (a) - Right Side
Confirm Balance for Equation (a)
Count Atoms in Equation (b) - Left Side
Count Atoms in Equation (b) - Right Side
Confirm Balance for Equation (b)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Stoichiometry
Think of balancing equations like following a recipe. If you alter one ingredient, you must adjust the others to keep the end dish perfect. Similarly, in chemistry, we adjust coefficients in a balanced equation to ensure the starting substances (reactants) transform perfectly into the end substances (products) without any elements left behind or lost.
Key points about stoichiometry include:
- Determines the proportions of elements involved in reactions.
- Uses coefficients to relate the quantities of substances.
- Ensures matter is conserved during reactions.
Chemical Reactions
Examples from the exercise include the combustion reactions of hydrocarbons like propene (\(\mathrm{C}_{3}\mathrm{H}_{8}\)) and octane (\(\mathrm{C}_{8}\mathrm{H}_{18}\)). These involve hydrocarbons reacting with oxygen to form carbon monoxide (\(\mathrm{CO}\)), carbon dioxide (\(\mathrm{CO}_2\)), and water (\(\mathrm{H}_2\mathrm{O}\)). Such combustion reactions are common sources of energy.
Key features of chemical reactions:
- Change in substance composition.
- Energy exchange, often in the form of heat or light.
- Formation of new products from reactants.
Conservation of Mass
In our exercise, counting atoms on each side of the equation confirms this principle. For example, in both equations provided:
- The number of carbon atoms remains the same before and after the reaction.
- The number of hydrogen atoms are equal on both sides.
- The total number of oxygen atoms is consistent as well.