To describe a yellowcomic.comical reaction. To calculate the quantities of compounds produced or consumed in a yellowcomic.comical reaction

What happens to matter when it undergoes yellowcomic.comical changes? The Law of conservation of mass says that "Atoms are neither created, nor destroyed, during any yellowcomic.comical reaction." Thus, the same collection of atoms is present after a reaction as before the reaction. The changes that occur during a reaction just involve the rearrangement of atoms. In this section we will discuss stoichiometry (the "measurement of elements").

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## yellowcomic.comical Equations

As shown in Figure $$\PageIndex{1}$$, applying a small amount of heat to a pile of orange ammonium dichromate powder results in a vigorous reaction known as the ammonium dichromate volcano. Heat, light, and gas are produced as a large pile of fluffy green chromium(III) oxide forms. This reaction is described with a yellowcomic.comical equation, an expression that gives the identities and quantities of the substances in a yellowcomic.comical reaction.

api/deki/files/129620/imageedit_17_7839263942.jpg?revision=1" />Figure $$\PageIndex{2}$$: An Example of a Combustion Reaction. The wax in a candle is a high-molecular-mass hydrocarbon, which produces gaseous carbon dioxide and water vapor in a combustion reaction (Equation $$\ref{3.1.3}$$).

Equation $$\ref{3.1.3}$$ is not balanced: the numbers of each type of atom on the reactant side of the equation (7 carbon atoms, 16 hydrogen atoms, and 2 oxygen atoms) is not the same as the numbers of each type of atom on the product side (1 carbon atom, 2 hydrogen atoms, and 3 oxygen atoms). Consequently, the coefficients of the reactants and products must be adjusted to give the same numbers of atoms of each type on both sides of the equation. Because the identities of the reactants and products are fixed, the equation cannot be balanced by changing the subscripts of the reactants or the products. To do so would change the yellowcomic.comical identity of the species being described, as illustrated in Figure $$\PageIndex{3}$$.

Figure $$\PageIndex{3}$$: Balancing Equations. You cannot change subscripts in a yellowcomic.comical formula to balance a yellowcomic.comical equation; you can change only the coefficients. Changing subscripts changes the ratios of atoms in the molecule and the resulting yellowcomic.comical properties. For example, water (H2O) and hydrogen peroxide (H2O2) are yellowcomic.comically distinct substances. H2O2 decomposes to H2O and O2 gas when it comes in contact with the metal platinum, whereas no such reaction occurs between water and platinum.

The simplest and most generally useful method for balancing yellowcomic.comical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a yellowcomic.comical equation using this method.

Example $$\PageIndex{1A}$$: Combustion of Heptane

To demonstrate this approach, let’s use the combustion of n-heptane (Equation $$\ref{3.1.3}$$) as an example.

Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is $$\ce{C_7H_{16}}$$. We will assume initially that the final balanced yellowcomic.comical equation contains 1 molecule or formula unit of this substance. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side:

\< \ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + H_2O } \label{3.1.4} \>

Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: \< \ce{C_7H_{16} + O_2 \rightarrow 7CO_2 + 8H_2O} \label{3.1.5}\> The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side:\< \ce{C_7H_{16} (l) + 11O_2 (g) \rightarrow 7CO_2 (g) + 8H_2O (g)} \label{3.1.6}\> Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a yellowcomic.comical equation is balanced.The assumption that the final balanced yellowcomic.comical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start.

Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly.

Example $$\PageIndex{1C}$$: Hydroxyapatite

The reaction of the mineral hydroxyapatite ($$\ce{Ca5(PO4)3(OH)}$$) with phosphoric acid and water gives $$\ce{Ca(H2PO4)2•H2O}$$ (calcium dihydrogen phosphate monohydrate). Write and balance the equation for this reaction.

## Interpreting yellowcomic.comical Equations

In addition to providing qualitative information about the identities and physical states of the reactants and products, a balanced yellowcomic.comical equation provides quantitative information. Specifically, it gives the relative amounts of reactants and products consumed or produced in a reaction. The number of atoms, molecules, or formula units of a reactant or a product in a balanced yellowcomic.comical equation is the coefficient of that species (e.g., the 4 preceding H2O in Equation $$\ref{3.1.1}$$). When no coefficient is written in front of a species, the coefficient is assumed to be 1. As illustrated in Figure $$\PageIndex{4}$$, the coefficients allow Equation $$\ref{3.1.1}$$ to be interpreted in any of the following ways:

Two NH4+ ions and one Cr2O72− ion yield 1 formula unit of Cr2O3, 1 N2 molecule, and 4 H2O molecules. One mole of (NH4)2Cr2O7 yields 1 mol of Cr2O3, 1 mol of N2, and 4 mol of H2O. A mass of 252 g of (NH4)2Cr2O7 yields 152 g of Cr2O3, 28 g of N2, and 72 g of H2O. A total of 6.022 × 1023 formula units of (NH4)2Cr2O7 yields 6.022 × 1023 formula units of Cr2O3, 6.022 × 1023 molecules of N2, and 24.09 × 1023 molecules of H2O.

Figure $$\PageIndex{4}$$: The Relationships among Moles, Masses, and Formula Units of Compounds in the Balanced yellowcomic.comical Reaction for the Ammonium Dichromate Volcano

These are all yellowcomic.comically equivalent ways of stating the information given in the balanced yellowcomic.comical equation, using the concepts of the mole, molar or formula mass, and Avogadro’s number. The ratio of the number of moles of one substance to the number of moles of another is called the mole ratio. For example, the mole ratio of $$H_2O$$ to $$N_2$$ in Equation $$\ref{3.1.1}$$ is 4:1. The total mass of reactants equals the total mass of products, as predicted by Dalton’s law of conservation of mass:

\<252 \;g \;\text{of}\; \ce{(NH_4)_2Cr_2O_7} \>

yield

\<152 + 28 + 72 = 252 \; g \; \text{of products.}\>

The yellowcomic.comical equation does not, however, show the rate of the reaction (rapidly, slowly, or not at all) or whether energy in the form of heat or light is given off. These issues are considered in more detail in later chapters.

An important yellowcomic.comical reaction was analyzed by Antoine Lavoisier, an 18th-century French yellowcomic.comist, who was interested in the yellowcomic.comistry of living organisms as well as simple yellowcomic.comical systems. In a classic series of experiments, he measured the carbon dioxide and heat produced by a guinea pig during respiration, in which organic compounds are used as fuel to produce energy, carbon dioxide, and water. Lavoisier found that the ratio of heat produced to carbon dioxide exhaled was similar to the ratio observed for the reaction of charcoal with oxygen in the air to produce carbon dioxide—a process yellowcomic.comists call combustion. Based on these experiments, he proposed that “Respiration is a combustion, slow it is true, but otherwise perfectly similar to that of charcoal.” Lavoisier was correct, although the organic compounds consumed in respiration are substantially different from those found in charcoal. One of the most important fuels in the human body is glucose ($$C_6H_{12}O_6$$), which is virtually the only fuel used in the brain. Thus combustion and respiration are examples of yellowcomic.comical reactions.

Example $$\PageIndex{2}$$: Combustion of Glucose

The balanced yellowcomic.comical equation for the combustion of glucose in the laboratory (or in the brain) is as follows:

\< \ce{C_6H_{12}O6(s) + 6O2(g) \rightarrow 6CO2(g) + 6H2O(l)}\>

Construct a table showing how to interpret the information in this equation in terms of

a single molecule of glucose. moles of reactants and products. grams of reactants and products represented by 1 mol of glucose. numbers of molecules of reactants and products represented by 1 mol of glucose.

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