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Balance each of the following chemical equations. a. \(\mathrm{C}_{2}(g)+\mathrm{KBr}(a q) \rightarrow \mathrm{Br}_{2}(I)+\mathrm{KCl}(a q)\) b. \(\operatorname{Cr}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Cr}_{2} \mathrm{O}_{3}(s)\) c. \(P_{4}(s)+H_{2}(g) \rightarrow P H_{3}(g)\) d. \(\mathrm{Al}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+\mathrm{H}_{2}(g)\) e. \(\mathrm{PCl}_{3}(l)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{3} \mathrm{PO}_{3}(a q)+\mathrm{HCl}(a q)\) f. \(\mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow \mathrm{SO}_{3}(g)\) g. \(\mathrm{C}_{7} \mathrm{H}_{16}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) h. \(C_{2} H_{6}(g)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(g)\)

Short Answer

Expert verified
a. C鈧(g) + 2 KBr(aq) 鈫 Br鈧(l) + 2 KCl(aq) b. 2 Cr(s) + O鈧(g) 鈫 Cr鈧侽鈧(s) c. P鈧(s) + 6 H鈧(g) 鈫 4 PH鈧(g) d. 2 Al(s) + 3 H鈧係O鈧(aq) 鈫 Al鈧(SO鈧)鈧(aq) + 3 H鈧(g) e. PCl鈧(l) + 3 H鈧侽(l) 鈫 H鈧働O鈧(aq) + 3 HCl(aq) f. 2 SO鈧(g) + O鈧(g) 鈫 2 SO鈧(g) g. C鈧嘓鈧佲倖(l) + 11 O鈧(g) 鈫 7 CO鈧(g) + 8 H鈧侽(g) h. 2 C鈧侶鈧(g) + 7 O鈧(g) 鈫 4 CO鈧(g) + 6 H鈧侽(g)

Step by step solution

01

a. Balancing C鈧(g) + KBr(aq) 鈫 Br鈧(l) + KCl(aq)

We start by listing the number of atoms of each element on both sides of the equation: - Carbon: 2 on the left, 0 on the right - Potassium: 1 on the left, 1 on the right - Bromine: 1 on the left, 2 on the right - Chlorine: 0 on the left, 1 on the right To balance the equation: 1. Add a coefficient of 2 in front of KCl: C鈧(g) + KBr(aq) 鈫 Br鈧(l) + 2 KCl(aq) 2. Add a coefficient of 2 in front of KBr: C鈧(g) + 2 KBr(aq) 鈫 Br鈧(l) + 2 KCl(aq) Now the equation is balanced.
02

b. Balancing Cr(s) + O鈧(g) 鈫 Cr鈧侽鈧(s)

To balance the equation: 1. Add a coefficient of 2 in front of Cr: 2 Cr(s) + O鈧(g) 鈫 Cr鈧侽鈧(s) Now the equation is balanced.
03

c. Balancing P鈧(s) + H鈧(g) 鈫 PH鈧(g)

To balance the equation: 1. Add a coefficient of 4 in front of PH鈧: P鈧(s) + H鈧(g) 鈫 4 PH鈧(g) 2. Add a coefficient of 6 in front of H鈧: P鈧(s) + 6 H鈧(g) 鈫 4 PH鈧(g) Now the equation is balanced.
04

d. Balancing Al(s) + H鈧係O鈧(aq) 鈫 Al鈧(SO鈧)鈧(aq) + H鈧(g)

To balance the equation: 1. Add a coefficient of 2 in front of Al: 2 Al(s) + H鈧係O鈧(aq) 鈫 Al鈧(SO鈧)鈧(aq) + H鈧(g) 2. Add a coefficient of 3 in front of H鈧係O鈧: 2 Al(s) + 3 H鈧係O鈧(aq) 鈫 Al鈧(SO鈧)鈧(aq) + H鈧(g) 3. Add a coefficient of 3 in front of H鈧: 2 Al(s) + 3 H鈧係O鈧(aq) 鈫 Al鈧(SO鈧)鈧(aq) + 3 H鈧(g) Now the equation is balanced.
05

e. Balancing PCl鈧(l) + H鈧侽(l) 鈫 H鈧働O鈧(aq) + HCl(aq)

To balance the equation: 1. Add a coefficient of 3 in front of HCl: PCl鈧(l) + H鈧侽(l) 鈫 H鈧働O鈧(aq) + 3 HCl(aq) 2. Add a coefficient of 3 in front of H鈧侽: PCl鈧(l) + 3 H鈧侽(l) 鈫 H鈧働O鈧(aq) + 3 HCl(aq) Now the equation is balanced.
06

f. Balancing SO鈧(g) + O鈧(g) 鈫 SO鈧(g)

To balance the equation: 1. Add a coefficient of 2 in front of SO鈧: SO鈧(g) + O鈧(g) 鈫 2 SO鈧(g) 2. Add a coefficient of 2 in front of SO鈧: 2 SO鈧(g) + O鈧(g) 鈫 2 SO鈧(g) Now the equation is balanced.
07

g. Balancing C鈧嘓鈧佲倖(l) + O鈧(g) 鈫 CO鈧(g) + H鈧侽(g)

To balance the equation: 1. Add a coefficient of 7 in front of CO鈧: C鈧嘓鈧佲倖(l) + O鈧(g) 鈫 7 CO鈧(g) + H鈧侽(g) 2. Add a coefficient of 8 in front of H鈧侽: C鈧嘓鈧佲倖(l) + O鈧(g) 鈫 7 CO鈧(g) + 8 H鈧侽(g) 3. Add a coefficient of 11 in front of O鈧: C鈧嘓鈧佲倖(l) + 11 O鈧(g) 鈫 7 CO鈧(g) + 8 H鈧侽(g) Now the equation is balanced.
08

h. Balancing C鈧侶鈧(g) + O鈧(g) 鈫 CO鈧(g) + H鈧侽(g)

To balance the equation: 1. Add a coefficient of 2 in front of CO鈧: C鈧侶鈧(g) + O鈧(g) 鈫 2 CO鈧(g) + H鈧侽(g) 2. Add a coefficient of 3 in front of H鈧侽: C鈧侶鈧(g) + O鈧(g) 鈫 2 CO鈧(g) + 3 H鈧侽(g) 3. Add a coefficient of 7/2 in front of O鈧: C鈧侶鈧(g) + 7/2 O鈧(g) 鈫 2 CO鈧(g) + 3 H鈧侽(g) Alternatively, to avoid fractional coefficients, we can multiply the entire equation by 2: C鈧侶鈧(g) + 3.5 O鈧(g) 鈫 2 CO鈧(g) + 3 H鈧侽(g) becomes 2 C鈧侶鈧(g) + 7 O鈧(g) 鈫 4 CO鈧(g) + 6 H鈧侽(g) Now the equation is balanced.

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Key Concepts

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

Stoichiometry
Stoichiometry is an essential concept in chemistry, focusing on the relationship between the quantities of reactants and products in a chemical reaction. When balancing chemical equations, stoichiometry ensures that the number of atoms for each element is the same on both sides of the equation. This is crucial because it maintains the integrity of the chemical process and enables us to predict the quantities needed and those that will be produced.

In order to apply stoichiometry effectively, you must first comprehend the coefficients in a balanced chemical equation. These coefficients indicate the proportion of moles, molecules, or volumes in which the substances react. Understanding these proportions lets us use stoichiometry to solve real-world problems, such as calculating how much of a reactant is needed to produce a certain amount of product.

Quick facts about stoichiometry:
  • It utilizes ratios derived from chemical equations.
  • It allows conversion between moles, grams, and particles.
  • Helps determine limiting reactants and theoretical yield.
  • Fundamental for quantitative chemical analysis.
Chemical Reactions
Chemical reactions involve the breaking of bonds in reactants and the formation of new bonds to create products. A balanced chemical equation represents this transformation. Different types of chemical reactions, such as synthesis, decomposition, single replacement, and combustion, have unique characteristics and patterns.

In a balanced equation, both the type and number of atoms remain constant between reactants and products, which is a fundamental characteristic of all chemical reactions. Understanding the different types of reactions helps predict product formation and energy changes. For example:
  • Combustion: A reaction where a substance combines with oxygen, releasing heat and often forming carbon dioxide and water, such as the burning of hydrocarbons like ethane or hexane.
  • Synthesis: Two or more simple substances combine to form a more complex product, such as forming aluminum sulfate from aluminum and sulfuric acid.
  • Decomposition: A complex molecule breaks down into simpler products, often involving heating or other energy input.

Recognizing how reactants transform into products allows you to construct and balance chemical reactions effectively and accurately.
Law of Conservation of Mass
The Law of Conservation of Mass states that mass is conserved in a chemical reaction; it cannot be created or destroyed. This law is the foundation for balancing chemical equations. It implies that the total mass of reactants equals the total mass of products. Therefore, all atoms present in reactants must be accounted for in the products.

In practical terms, when balancing a chemical equation, each type of atom must appear in the same quantity on both sides of the equation. If any discrepancies exist, the equation is unbalanced, indicating a violation of this fundamental law. For example, as seen in the balanced equation of a combustion reaction like C鈧侶鈧 and O鈧 forming CO鈧 and H鈧侽, each carbon and oxygen atom present in the reactants corresponds exactly to those in the products.

Balancing a chemical equation also honors this principle, ensuring that all matter is accounted for through proper stoichiometric coefficients. This process validates our calculations and predictions about how substances will behave and combine during reactions.
Balancing Methods
Balancing chemical equations is a skill that involves several methods to ensure all elements have the same number of atoms on both sides of the reaction. Here are some common techniques for balancing equations efficiently:
  • Inspection Method: This involves adding coefficients to adjust the number of atoms of each element, starting with one that appears in a single reactant and product. Adjust larger, more complex molecules last.
  • Algebraic Method: Helps in balancing complex reactions by setting up algebraic equations based on the number of each type of atom.
  • Oxidation Number Method: Often used for redox reactions, it focuses on the change in oxidation states to achieve balance.
Regardless of the method, the goal is to balance the equation so that it satisfies the Law of Conservation of Mass. During this process, keep in mind:
  • You can only adjust coefficients (numbers in front of molecules), not subscripts within formulas.
  • Double-check atom numbers for populous elements in reactants and products.
  • Balancing coefficients should be the smallest set of integers possible.
Understanding these processes through practice will make balancing chemical equations straightforward and intuitive.

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Most popular questions from this chapter

When a strip of magnesium metal is heated in oxygen, it bursts into an intensely white flame and produces a finely powdered dust of magnesium oxide. Write the unbalanced chemical equation for this process.

Elemental boron is produced in one industrial process by heating diboron trioxide with magnesium metal, also producing magnesium oxide as a by-product. Write the unbalanced chemical equation for this process.

Balance each of the following chemical equations. a. \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{K}_{2} \mathrm{CrO}_{4}(a q) \rightarrow\) \(\mathrm{PbCrO}_{4}(s)+\mathrm{KNO}_{3}(a q)\) b. \(\mathrm{BaCl}_{2}(a q)+\mathrm{Na}_{2} \mathrm{SO}_{4}(a q) \rightarrow \mathrm{BaSO}_{4}(s)+\mathrm{NaCl}(a q)\) c. \(\mathrm{CH}_{3} \mathrm{OH}(l)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{Na}_{2} \mathrm{CO}_{3}(a q)+\mathrm{S}(s)+\mathrm{SO}_{2}(g) \rightarrow\) \(\mathrm{CO}_{2}(g)+\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q)\) e. \(\mathrm{Cu}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \rightarrow\) \(\operatorname{CuSO}_{4}(a q)+\mathrm{SO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) f. \(\operatorname{MnO}_{2}(s)+\operatorname{HCl}(a q) \rightarrow \operatorname{MnCl}_{2}(a q)+\mathrm{Cl}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) g. \(\mathrm{As}_{2} \mathrm{O}_{3}(s)+\mathrm{KI}(a q)+\mathrm{HCl}(a q) \rightarrow\) \(\mathrm{AsI}_{3}(s)+\mathrm{KCl}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) h. \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3}(a q)+\mathrm{I}_{2}(a q) \rightarrow \mathrm{Na}_{2} \mathrm{S}_{4} \mathrm{O}_{6}(a q)+\mathrm{NaI}(a q)\)

Balance each of the following chemical equations. a. \(\mathrm{Li}(s)+\mathrm{Cl}_{2}(g) \rightarrow \mathrm{LiCl}(s)\) b. \(\mathrm{Ba}(s)+\mathrm{N}_{2}(g) \rightarrow \mathrm{Ba}_{3} \mathrm{N}_{2}(s)\) c. \(\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{Al}(s)+\mathrm{HCl}(a q) \rightarrow \mathrm{AlCl}_{3}(a q)+\mathrm{H}_{2}(g)\) e. \(\operatorname{NiS}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{NiO}(s)+\mathrm{SO}_{2}(g)\) f. \(\operatorname{CaH}_{2}(s)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{Ca}(\mathrm{OH})_{2}(s)+\mathrm{H}_{2}(g)\) g. \(\mathrm{H}_{2}(g)+\mathrm{CO}(g) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(l)\) h. \(\mathrm{B}_{2} \mathrm{O}_{3}(s)+\mathrm{C}(s) \rightarrow \mathrm{B}_{4} \mathrm{C}_{3}(s)+\mathrm{CO}_{2}(g)\)

After balancing a chemical equation, we ordinarily make sure that the coefficients are the smallest ________ possible.

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