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Chapter 4: Problem 15

Complete and balance each combustion equation. a) \(\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}+\mathrm{O}_{2} \rightarrow\) ? b) \(\mathrm{CH}_{3} \mathrm{NO}_{2}+\mathrm{O}_{2} \rightarrow ?\)

Short Answer

Expert verified
a) \( 2\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} + 13\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 10\mathrm{H}_2O \). b) \( 2\mathrm{CH}_3 \mathrm{NO}_2 + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2O + \mathrm{N}_2 \).

Step by step solution

01

Understand Combustion Reaction

Combustion reactions involve a hydrocarbon (or related compound) reacting with oxygen to produce carbon dioxide and water. The general formula for a combustion reaction is: \[ \text{Hydrocarbon} + O_2 \rightarrow CO_2 + H_2O \]
02

Write Unbalanced Equation for A

For the compound \( \mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} \), which is butanol, we write the unbalanced equation: \[ \mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2O \]
03

Balance Equation for A

To balance the equation, start by balancing the carbon atoms. There are 4 carbon atoms in \( \mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} \), so we need 4 \( \mathrm{CO}_2 \): \[ \mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} + \mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + \mathrm{H}_2O \] Next, balance the hydrogen atoms. There are 10 hydrogen atoms in \( \mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} \), so we need 5 \( \mathrm{H}_2O \): \[ \mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} + \mathrm{O}_2 \rightarrow 4\mathrm{CO}_2 + 5\mathrm{H}_2O \] Now, balance the oxygen atoms. There are a total of 13 oxygen atoms needed (8 from \( 4 \mathrm{CO}_2 \) and 5 from \( 5 \mathrm{H}_2O \)), and since \( \mathrm{O}_2 \) is a diatomic molecule, we need \( 13/2 \) or 6.5 \( \mathrm{O}_2 \). To avoid fractions, multiply the entire equation by 2: \[ 2\mathrm{C}_4 \mathrm{H}_9 \mathrm{OH} + 13\mathrm{O}_2 \rightarrow 8\mathrm{CO}_2 + 10\mathrm{H}_2O \]
04

Write Unbalanced Equation for B

For the compound \( \mathrm{CH}_3 \mathrm{NO}_2 \), we write the unbalanced equation: \[ \mathrm{CH}_3 \mathrm{NO}_2 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2O + \mathrm{N}_2 \] Note: In combustion, nitrogen usually forms \( \mathrm{N}_2 \) gas.
05

Balance Equation for B

To balance the equation, start by balancing the carbon atoms. There is 1 carbon atom in \( \mathrm{CH}_3 \mathrm{NO}_2 \), so we keep 1 \( \mathrm{CO}_2 \): \[ \mathrm{CH}_3 \mathrm{NO}_2 + \mathrm{O}_2 \rightarrow \mathrm{CO}_2 + \mathrm{H}_2O + \mathrm{N}_2 \] Next, balance the hydrogen atoms. There are 3 hydrogen atoms in \( \mathrm{CH}_3 \mathrm{NO}_2 \), so we need \( 1.5 \mathrm{H}_2O \). To avoid fractions initially: \[ 2\mathrm{CH}_3 \mathrm{NO}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2O + \mathrm{N}_2 \] Now, balance nitrogen. There are 2 nitrogen atoms in 2 molecules of \( \mathrm{CH}_3 \mathrm{NO}_2 \), so we have \( \mathrm{N}_2 \): \[ 2\mathrm{CH}_3 \mathrm{NO}_2 + \mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2O + \mathrm{N}_2 \] Finally, balance the oxygen atoms. We need 7 oxygen atoms (4 from 2 \( \mathrm{CO}_2 \), 3 from 3 \( \mathrm{H}_2O \)). Since each \( \mathrm{CH}_3 \mathrm{NO}_2 \) contributes 2 oxygen atoms, we need 3 more from \( \mathrm{O}_2 \), or 1.5 \( \mathrm{O}_2 \) per molecule of \( \mathrm{CH}_3 \mathrm{NO}_2 \), leading to: \[ 3\mathrm{O}_2 \] Thus, the balanced equation is: \[ 2\mathrm{CH}_3 \mathrm{NO}_2 + 3\mathrm{O}_2 \rightarrow 2\mathrm{CO}_2 + 3\mathrm{H}_2O + \mathrm{N}_2 \]

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Key Concepts

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

Chemical Equation Balancing
Balancing chemical equations is a fundamental skill in chemistry. It ensures that the law of conservation of mass is obeyed鈥攎eaning matter is neither created nor destroyed in a reaction. Every chemical equation represents a reaction, and balancing ensures that the number of each type of atom on the reactant side equals the number on the product side. Here's how it works:
  • Identify each type of atom present in the reactants and products.
  • Count how many atoms of each type you have on both sides.
  • Add coefficients (whole numbers placed before compounds) to ensure equal numbers of each atom on both sides.
  • Check your work by recounting atoms to confirm balance.
During this process, it's crucial to only adjust the coefficients, not the subscripts within compounds. This maintains the integrity of the chemical structure. When balanced correctly, the equation not only conserves mass but also tells us the proportions in which substances react and are formed.
Hydrocarbon Combustion
Hydrocarbon combustion is a specific type of chemical reaction where a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water. These reactions are a vital part of energy production, as seen in engines and power plants. The basic form for any combustion of hydrocarbon can be generalized as:
\[ \text{Hydrocarbon} + O_2 \rightarrow CO_2 + H_2O \]
Key points to understand this reaction include:
  • "Hydrocarbons" consist mainly of carbon (C) and hydrogen (H), often accompanied by other elements like oxygen in alcohols (e.g., butanol).
  • The reaction with O\(_2\) is highly exothermic, meaning it releases energy, often in the form of heat and light.
  • For a complete combustion, enough oxygen must be present to ensure that the hydrocarbons are entirely converted into CO\(_2\) and water.
Hence, the challenge lies in balancing the equation to reflect complete conversion, which involves aligning the number of C, H, and O atoms on both sides.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. When a chemical equation is balanced, stoichiometry allows us to use the coefficients to determine proportions. This means we can calculate precisely how much of each reactant is needed to produce a desired amount of product. Here's how to apply it:
  • Use the balanced equation to determine the mole ratio between reactants and products.
  • Convert masses or volumes of substances into moles, using their molecular weights.
  • Apply the mole ratios to calculate unknown quantities, whether it's amount of a reactant needed or the amount of product formed.
For instance, in the balanced butanol combustion equation provided, the ratio of butanol to oxygen is 2:13, meaning for every 2 moles of butanol, 13 moles of oxygen are needed. Stoichiometry helps convert these ratios into practical laboratory terms; it's pivotal in both understanding and predicting chemical reaction outcomes.

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Which is a combustion reaction and which is not? a) \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{O}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{O}_{2}\) b) \(\mathrm{C}_{2} \mathrm{H}_{4}+\mathrm{Cl}_{2} \rightarrow \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}\)
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