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Magazine Chapter 19: Problem 4
Balance the following equations for combustion reactions. Circle the fuel for each reaction. a. \(C_{2} H_{6}(g)+O_{2}(g) \longrightarrow CO_{2}(g)+H_{2} O(l)\) b. \(C_{6} H_{12} O_{6}(s)+O_{2}(g) \longrightarrow CO_{2}(g)+H_{2} O(l)\) c. \(C_{2} H_{5} OH(l)+O_{2}(g) \longrightarrow CO_{2}(g)+H_{2} O(l)\) d. \(C_{21} H_{24} N_{2} O_{4}(s)+O_{2}(g) \longrightarrow CO_{2}(g)+H_{2} O(l)+NO_{2}(g)\) e. \(C_{2} H_{5} SH(l)+O_{2}(g) \longrightarrow CO_{2}(g)+H_{2} O(l)+SO_{2}(g)\)
Short Answer
Expert verified
a. Fuel: \( C_{2}H_{6} \), b. Fuel: \( C_{6}H_{12}O_{6} \), c. Fuel: \( C_{2}H_{5}OH \), d. Fuel: \( C_{21}H_{24}N_{2}O_{4} \), e. Fuel: \( C_{2}H_{5}SH \).
Step by step solution
01
Balance Equation (a)
Start with equation (a) for the combustion of ethane: \( C_{2}H_{6} + O_{2} \rightarrow CO_{2} + H_{2}O \). First, balance carbon by placing a 2 in front of \( CO_{2} \): \( C_{2}H_{6} + O_{2} \rightarrow 2CO_{2} + H_{2}O \). Next, balance hydrogen by placing a 3 in front of \( H_{2}O \): \( C_{2}H_{6} + O_{2} \rightarrow 2CO_{2} + 3H_{2}O \). Finally, balance oxygen by placing a 7/2 in front of \( O_{2} \): \( C_{2}H_{6} + \frac{7}{2}O_{2} \rightarrow 2CO_{2} + 3H_{2}O \). To clear the fraction, multiply the entire equation by 2: \( 2C_{2}H_{6} + 7O_{2} \rightarrow 4CO_{2} + 6H_{2}O \). The fuel in this reaction is \( C_{2}H_{6} \).
02
Balance Equation (b)
Next, consider equation (b) for glucose: \( C_{6}H_{12}O_{6} + O_{2} \rightarrow CO_{2} + H_{2}O \). Balance carbons by placing a 6 in front of \( CO_{2} \): \( C_{6}H_{12}O_{6} + O_{2} \rightarrow 6CO_{2} + H_{2}O \). Balance hydrogen by placing a 6 in front of \( H_{2}O \): \( C_{6}H_{12}O_{6} + O_{2} \rightarrow 6CO_{2} + 6H_{2}O \). Balance oxygen by placing 6 in front of \( O_{2} \): \( C_{6}H_{12}O_{6} + 6O_{2} \rightarrow 6CO_{2} + 6H_{2}O \). The fuel is \( C_{6}H_{12}O_{6} \).
03
Balance Equation (c)
For the combustion of ethanol \( C_{2}H_{5}OH \): \( C_{2}H_{5}OH + O_{2} \rightarrow CO_{2} + H_{2}O \). Balance carbon by placing a 2 in front of \( CO_{2} \): \( C_{2}H_{5}OH + O_{2} \rightarrow 2CO_{2} + H_{2}O \). Balance hydrogen by placing a 3 in front of \( H_{2}O \): \( C_{2}H_{5}OH + O_{2} \rightarrow 2CO_{2} + 3H_{2}O \). Balance oxygen by recognizing that you need 7 oxygen atoms on the right, leading to \( C_{2}H_{5}OH + 3O_{2} \rightarrow 2CO_{2} + 3H_{2}O \). The fuel is \( C_{2}H_{5}OH \).
04
Balance Equation (d)
Consider equation (d) for the combustion of caffeine: \( C_{21}H_{24}N_{2}O_{4} + O_{2} \rightarrow CO_{2} + H_{2}O + NO_{2} \). Start by balancing carbon with a 21 in front of \( CO_{2} \): \( C_{21}H_{24}N_{2}O_{4} + O_{2} \rightarrow 21CO_{2} + H_{2}O + NO_{2} \). Balance hydrogen with a 12 in front of \( H_{2}O \): \( C_{21}H_{24}N_{2}O_{4} + O_{2} \rightarrow 21CO_{2} + 12H_{2}O + NO_{2} \). Balance nitrogen with a 2 in front of \( NO_{2} \): \( C_{21}H_{24}N_{2}O_{4} + O_{2} \rightarrow 21CO_{2} + 12H_{2}O + 2NO_{2} \). Balance the oxygen, leading to: \( C_{21}H_{24}N_{2}O_{4} + 25O_{2} \rightarrow 21CO_{2} + 12H_{2}O + 2NO_{2} \). The fuel is \( C_{21}H_{24}N_{2}O_{4} \).
05
Balance Equation (e)
Lastly, for the combustion of ethanethiol \( C_{2}H_{5}SH \): \( C_{2}H_{5}SH + O_{2} \rightarrow CO_{2} + H_{2}O + SO_{2} \). Balance carbon with a 2 in front of \( CO_{2} \): \( C_{2}H_{5}SH + O_{2} \rightarrow 2CO_{2} + H_{2}O + SO_{2} \). Balance hydrogen with a 3 in front of \( H_{2}O \): \( C_{2}H_{5}SH + O_{2} \rightarrow 2CO_{2} + 3H_{2}O + SO_{2} \). For sulfur, the equation already includes 1 \( SO_{2} \). Balance oxygen, yielding: \( C_{2}H_{5}SH + \frac{9}{2}O_{2} \rightarrow 2CO_{2} + 3H_{2}O + SO_{2} \). Multiply the entire equation by 2: \( 2C_{2}H_{5}SH + 9O_{2} \rightarrow 4CO_{2} + 6H_{2}O + 2SO_{2} \). The fuel is \( C_{2}H_{5}SH \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Combustion Reactions
Combustion reactions are fascinating chemical processes where a fuel reacts with an oxidizing agent, typically oxygen, releasing energy in the form of heat and light. This process is fundamental in many areas from energy production to propulsion.
In the simplest terms, imagine lighting a candle. The wax serves as the fuel that burns in the presence of oxygen (from air), producing a flame, which is the observable product of a combustion reaction.
In the simplest terms, imagine lighting a candle. The wax serves as the fuel that burns in the presence of oxygen (from air), producing a flame, which is the observable product of a combustion reaction.
- **Fuel and Oxygen:** The fuel in a combustion reaction can be various substances, often carbon-based, such as hydrocarbons or alcohols. Oxygen is usually involved to oxidize the fuel.
- **Energy Release:** Energy is released when the fuel's chemical bonds break and reform into new compounds like carbon dioxide and water.
- **Types of Combustion:** Complete combustion occurs when the fuel burns in sufficient oxygen, yielding carbon dioxide and water. Incomplete combustion, however, happens when there's limited oxygen, often producing carbon monoxide, a hazardous gas.
Balancing Equations
Balancing chemical equations is like solving a puzzle. The goal is to ensure the number of atoms for each element is the same on both sides of the reaction arrow. This reflects the conservation of mass principle, where atoms are neither created nor destroyed.
To balance an equation, start by comparing the number of each type of atom present in the reactants and products, adjusting coefficients as necessary.
Here's a simple strategy to balance equations:
To balance an equation, start by comparing the number of each type of atom present in the reactants and products, adjusting coefficients as necessary.
Here's a simple strategy to balance equations:
- **List Elements:** Write out all elements involved in the reaction.
- **Count Atoms:** Tally the number of atoms for each element on both sides of the equation.
- **Adjust Coefficients:** Alter the coefficients (the numbers in front of molecules, not to be confused with subscripts) to balance the atoms for each element.
- **Check Oxygen Last:** Often, oxygen is abundant in reactions, so adjust this element at the end if needed.
- **Verify:** Double-check all elements to ensure balance.
Organic Chemistry
Organic chemistry is the branch of chemistry that studies the structure, properties, composition, reactions, and preparation of carbon-containing compounds. These include not only hydrocarbons but also compounds with various other elements like hydrogen, nitrogen, oxygen, sulfur, and phosphorus.
Organic compounds are everywhere鈥攆rom the medication we take and the food we eat, to the fibers in our clothes.
Here's a glance at how organic chemistry interacts with combustion reactions:
Organic compounds are everywhere鈥攆rom the medication we take and the food we eat, to the fibers in our clothes.
Here's a glance at how organic chemistry interacts with combustion reactions:
- **Hydrocarbons:** Many fuels like methane, ethane, and ethanol are hydrocarbons. They combust to produce carbon dioxide and water, a common reaction studied in organic chemistry.
- **Alcohols and Other Compounds:** Ethanol and methanol are examples of organic alcohols involved in combustion reactions. These types of reactions showcase the adaptability and reactivity of different organic compounds.
- **Environmental Impact:** Organic chemistry also evaluates the by-products of combustion, such as soot or other pollutants, understanding their effect on health and the environment.
Stochiometry
Stochiometry is a central concept in chemistry, encompassing the quantitative aspects of chemical reactions. It provides the calculations needed to measure reactants and products in a reaction.
Understanding stoichiometry involves several key steps and ideas:
Understanding stoichiometry involves several key steps and ideas:
- **Mole Concept:** Stoichiometry revolves around the mole, a unit that describes a definite number of atoms, ions, or molecules. One mole is equivalent to Avogadro's number, approximately \(6.02 \times 10^{23}\) of these entities.
- **Molecular Weight:** Knowing the molecular weight of molecules helps in converting mass to moles and vice versa. This is critical for calculating the amounts needed or produced in a reaction.
- **Balanced Equations:** The coefficients in a balanced chemical equation act as a mole ratio. This ratio is essential in converting from reactants to products.
- **Limiting Reagent:** The limiting reagent is the reactant that runs out first, limiting the amount of product formed in a reaction. Understanding which component is the limiting reagent helps in evaluating reaction efficiency.
