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Chapter 3: Problem 103

Balance the following equations representing combustion reactions: c. \(C_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. Fe \((s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) e. \(\mathrm{FeO}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\)

Short Answer

Expert verified
The balanced equations are: (c) \(C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(g)\) (d) \(4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)\) (e) \(2FeO(s)+ O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\)

Step by step solution

01

Balancing Equation (c)

Given the equation: \(C_{12}H_{22}O_{11}(s)+ O_{2}(g) \rightarrow CO_{2}(g)+ H_{2}O(g)\) 1. Count the atoms of each element on both sides of the equation. Left side: 12 C, 22 H, and 12 O atoms Right side: 1 C, 2 H, and 3 O atoms 2. Balance the Carbon atoms by adding a coefficient to the CO鈧� term on the right side: \(C_{12}H_{22}O_{11}(s) + O_{2}(g) \rightarrow 12CO_{2}(g) + H_{2}O(g)\) 3. Balance the Hydrogen atoms by adding a coefficient to the H鈧侽 term on the right side: \(C_{12}H_{22}O_{11}(s) + O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(g)\) 4. Finally, balance the Oxygen atoms by adding a coefficient to the O鈧� term on the left side: \(C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(g)\) The balanced equation for (c) is: \(C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(g)\)
02

Balancing Equation (d)

Given the equation: \(Fe(s) + O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\) 1. Count the atoms of each element on both sides of the equation. Left side: 1 Fe and 2 O atoms Right side: 2 Fe and 3 O atoms 2. Balance the Iron atoms by adding a coefficient to the Fe term on the left side: \(4Fe(s) + O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)\) 3. Balance the Oxygen atoms by adding a coefficient to the O鈧� term on the left side: \(4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)\) The balanced equation for (d) is: \(4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)\)
03

Balancing Equation (e)

Given the equation: \(FeO(s)+ O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\) 1. Count the atoms of each element on both sides of the equation. Left side: 1 Fe, and 3 O atoms Right side: 2 Fe and 3 O atoms 2. Balance the Iron atoms by adding a coefficient to the FeO term on the left side: \(2FeO(s)+ O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\) 3. Balance the Oxygen atoms (already balanced, no need for change): \(2FeO(s)+ O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\) The balanced equation for (e) is: \(2FeO(s)+ O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\)

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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 chemical processes in which a substance combines with oxygen, releasing energy in the form of heat or light. This process often results in the production of carbon dioxide and water, especially when organic compounds are involved. These reactions are exothermic, meaning they release energy. The most common example of a combustion reaction is the burning of fuel in an internal combustion engine. In a typical combustion reaction:
  • The reactants are a hydrocarbon and oxygen.
  • The products are carbon dioxide and water.
Understanding how to balance these reactions is crucial because it reflects the conservation of mass, indicating that matter cannot be created or destroyed in a chemical reaction. By accurately balancing combustion reactions, we can predict the amount of products formed and the energy released.
Stoichiometry
Stoichiometry is the area of chemistry that focuses on the quantitative relationships of reactants and products in a chemical reaction. It is like the baking recipe of science that tells us how much of each ingredient (reactant) we need and what we鈥檒l get in return (products). To apply stoichiometry effectively:
  • Start by writing a balanced chemical equation.
  • Use the coefficients from the balance equation to determine mole ratios between reactants and products.
  • Calculate the moles of substances, and convert them into grams if needed, using their molar masses.
This concept helps in understanding the efficiency of reactions, determining limiting reactants, and calculating theoretical yields. It is also essential for scaling reactions from lab settings to industrial production.
Chemical Reactions
A chemical reaction is a transformation where one or more substances (reactants) are changed into one or more new substances (products). It represents a rearrangement of atoms and involves breaking and forming chemical bonds. There are several types of chemical reactions, including:
  • Synthesis reactions - where compounds are built up.
  • Decomposition reactions - where compounds are broken down.
  • Single replacement reactions - where one element replaces another in a compound.
  • Double replacement reactions - where ions in two compounds swap.
  • Combustion reactions - typically involving oxygen and producing energy.
Recognizing the type of reaction occurring helps predict the products and the energy changes involved. It also aids in understanding the equilibrium state of a reaction and how to manipulate conditions to favor desired products.
Elemental Balancing
Elemental balancing is crucial for making sense of chemical reactions, ensuring that mass is conserved. It involves adjusting the coefficients (the numbers placed before molecules in an equation) so that the number of each atom type is the same on both sides of the equation. The process involves:
  • Counting the number of each type of atom in the reactants and products.
  • Adjusting coefficients methodically, often starting with the most complex molecule.
  • Ensuring all the atoms balance correctly, checking both sides continuously.
Balanced equations not only obey the Law of Conservation of Mass but also serve as the foundation for stoichiometric calculations. Without this step being accurate, any subsequent calculations about quantities needed or produced would be incorrect, affecting both laboratory and industrial practices.

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Most popular questions from this chapter

A substance \(\mathrm{X}_{2} \mathrm{Z}\) has the composition (by mass) of 40.0\% X and 60.0\(\% \mathrm{Z}\) . What is the composition (by mass) of the compound \(\mathrm{XZ}_{2} ?\)

Give the balanced equation for each of the following. a. The combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. b. Aqueous solutions of lead(Il) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of lead(II) phosphate with aqueous sodium nitrate as the other product. c. Solid zinc reacts with aqueous HCl to form aqueous zinc chloride and hydrogen gas. d. Aqueous strontium hydroxide reacts with aqueous hydrobromic acid to produce water and aqueous strontium bromide.

A compound contains 47.08\(\%\) carbon, 6.59\(\%\) hydrogen, and 46.33\(\%\) chlorine by mass; the molar mass of the compound is 153 g/mol. What are the empirical and molecular formulas of the compound?

The reusable booster rockets of the U.S. space shuttle employ a mixture of aluminum and ammonium perchlorate for fuel. A possible equation for this reaction is $$ 3 \mathrm{Al}(s)+3 \mathrm{NH}_{4} \mathrm{ClO}_{4}(s) \longrightarrow $$ $$ \mathrm{Al}_{2} \mathrm{O}_{3}(s)+\mathrm{AlCl}_{3}(s)+3 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ What mass of \(\mathrm{NH}_{4} \mathrm{ClO}_{4}\) should be used in the fuel mixture for every kilogram of Al?

A sample of a hydrocarbon (a compound consisting of only carbon and hydrogen contains \(2.59 \times 10^{23}\) atoms of hydrogen and is 17.3\(\%\) hydrogen by mass. If the molar mass of the hydrocarbon is between 55 and 65 g/mol, what amount (moles) of compound is present, and what is the mass of the sample?
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