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Chapter 3: Problem 28

A balanced chemical equation contains a large amount of information. What information is given in a balanced equation?

Short Answer

Expert verified
A balanced chemical equation provides information about the reactants and products involved in a chemical reaction, their stoichiometry (represented by coefficients), conservation of mass, and the state of each substance (e.g., solid, liquid, gas, aqueous solution). Additionally, some balanced equations may provide information about reaction conditions, such as temperature, pressure, or catalysts. This information is essential for understanding, predicting, and optimizing chemical reactions in various applications.

Step by step solution

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1. Basics of a Balanced Chemical Equation

A balanced chemical equation contains chemical formulas for the reactants and products, along with coefficients to denote the number of each substance. It represents a chemical reaction in which the number of atoms for each element in the reactants is equal to the number of atoms in the products. This follows the Law of Conservation of Mass, which states that matter cannot be created or destroyed.
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2. Components of a Balanced Chemical Equation

A balanced chemical equation includes: a. Reactants: The substances that react in the chemical equation. b. Products: The substances that are formed as a result of the reaction. c. Coefficients: The coefficients indicate the number of molecules or moles of a substance. They are whole numbers that are written in front of the chemical formulas to balance the equation.
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3. Information given by a Balanced Chemical Equation

A balanced equation provides the following information: a. Reactants and products: It describes the substances involved in a reaction and the substances that result from the reaction. b. Stoichiometry: The coefficients in a balanced chemical equation give information about the ratio of the reactants and products in a reaction. This information can help to determine the amount of a substance that will be consumed or produced in the reaction. c. Conservation of mass: The balanced equation demonstrates the Law of Conservation of Mass by showing that the total number of atoms in the reactants is equal to the total number of atoms in the products. d. State of reactants and products: The equation can also include labels to indicate the state of each substance, such as (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solutions. e. Reaction conditions: Some balanced chemical equations may also provide information about reaction conditions, such as temperature, pressure, or catalysts. These factors can affect the rate and extent of the reaction.
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4. Importance of Balanced Chemical Equations

Balanced chemical equations are essential for understanding and predicting chemical reactions because they provide crucial information about the reactants, products, stoichiometry, and conservation of mass. This information is useful for a range of applications, including laboratory experiments, industrial processes, and environmental studies. By interpreting the information given in balanced equations, scientists and engineers can design, control, and optimize chemical processes.

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Most popular questions from this chapter

Consider the following data for three binary compounds of hydrogen and nitrogen: $$ \begin{array}{lcc} & \% \mathrm{H} \text { (by Mass) } & \text { \% N (by Mass) } \\ \hline \text { I } & 17.75 & 82.25 \\ \text { II } & 12.58 & 87.42 \\ \text { III } & 2.34 & 97.66 \end{array} $$ When \(1.00 \mathrm{~L}\) of each gaseous compound is decomposed to its elements, the following volumes of \(\mathrm{H}_{2}(g)\) and \(\mathrm{N}_{2}(g)\) are obtained: $$ \begin{array}{lcc} & \mathrm{H}_{2} \text { (L) } & \mathrm{N}_{2} \text { (L) } \\ \hline \text { I } & 1.50 & 0.50 \\ \text { II } & 2.00 & 1.00 \\ \text { III } & 0.50 & 1.50 \end{array} $$ Use these data to determine the molecular formulas of compounds I, II, and III and to determine the relative values for the atomic masses of hydrogen and nitrogen.

The element rhenium (Re) has two naturally occurring isotopes, \({ }^{185} \mathrm{Re}\) and \({ }^{187} \mathrm{Re}\), with an average atomic mass of \(186.207\) amu. Rhenium is \(62.60 \%^{187} \mathrm{Re}\), and the atomic mass of \({ }^{187} \mathrm{Re}\) is \(186.956\) amu. Calculate the mass of \({ }^{185} \mathrm{Re}\).

Phosphorus can be prepared from calcium phosphate by the following reaction: \(2 \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)+6 \mathrm{SiO}_{2}(s)+10 \mathrm{C}(s) \longrightarrow\) $$ 6 \mathrm{CaSiO}_{3}(s)+\mathrm{P}_{4}(s)+10 \mathrm{CO}(g) $$ Phosphorite is a mineral that contains \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) plus other nonphosphorus- containing compounds. What is the maximum amount of \(\mathrm{P}_{4}\) that can be produced from \(1.0 \mathrm{~kg}\) of phosphorite if the phorphorite sample is \(75 \% \mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) by mass? Assume an excess of the other reactants.

Nitric acid is produced commercially by the Ostwald process, represented by the following equations: $$ \begin{aligned} 4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(g) & \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) & \longrightarrow 2 \mathrm{NO}_{2}(g) \\ 3 \mathrm{NO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l) & \longrightarrow 2 \mathrm{HNO}_{3}(a q)+\mathrm{NO}(g) \end{aligned} $$ What mass of \(\mathrm{NH}_{3}\) must be used to produce \(1.0 \times 10^{6} \mathrm{~kg} \mathrm{HNO}_{3}\) by the Ostwald process? Assume \(100 \%\) yield in each reaction and assume that the NO produced in the third step is not recycled.

Many homes in rural America are heated by propane gas, a compound that contains only carbon and hydrogen. Complete combustion of a sample of propane produced \(2.641 \mathrm{~g}\) of carbon dioxide and \(1.442 \mathrm{~g}\) of water as the only products. Find the empirical formula of propane.
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