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Chapter 3: Problem 20

Balance the following equations, and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (c) \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) (e) \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)\)

Short Answer

Expert verified
Reaction (a): \(\mathrm{C}_{3}\mathrm{H}_{6}(g)+\dfrac{9}{2}\mathrm{O}_{2}(g) \longrightarrow 3\mathrm{CO}_{2}(g)+3\mathrm{H}_{2}\mathrm{O}(g)\) - Combustion Reaction (b): \( \mathrm{NH}_{4}\mathrm{NO}_{3}(s) \longrightarrow 2\mathrm{N}_{2}\mathrm{O}(g)+2\mathrm{H}_{2}\mathrm{O}(g)\) - Decomposition Reaction (c): \(\mathrm{C}_{5}\mathrm{H}_{6}\mathrm{O}(l)+\dfrac{13}{2}\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g)+3\mathrm{H}_{2}\mathrm{O}(g)\) - Combustion Reaction (d): \(\mathrm{N}_{2}(g)+3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NH}_{3}(g)\) - Combination Reaction (e): \(\mathrm{K}_{2}\mathrm{O}(s)+\mathrm{H}_{2}\mathrm{O}(l) \longrightarrow 2\mathrm{KOH}(a q)\) - Combination

Step by step solution

01

Balancing Reaction (a)

Given reaction: \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Balance the elements in the following order: C, H, O. Balancing C: We have 3 C atoms on the reactant side, so we need 3 C atoms on the product side. \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow 3\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Balancing H: We have 6 H atoms on the reactant side, so we need 6 H atoms on the product side. \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\mathrm{O}_{2}(g) \longrightarrow 3\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\) Balancing O: Now we have 6 + 3 = 9 O atoms on the product side, so we need 9/2 O2 molecules on the reactant side. \(\mathrm{C}_{3} \mathrm{H}_{6}(g)+\dfrac{9}{2}\mathrm{O}_{2}(g) \longrightarrow 3\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\) Since there is oxygen gas as one of the reactants and the products are water and carbon dioxide, this reaction is a combustion reaction.
02

Balancing Reaction (b)

Given reaction: \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Balance the elements in the following order: N, H, O. Balancing N: We have 2 N atoms on the reactant side, so we need 2 N atoms on the product side. \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow 2\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Balancing H: We have 4 H atoms on the reactant side, so we need 4 H atoms on the product side. \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow 2\mathrm{N}_{2} \mathrm{O}(g)+2\mathrm{H}_{2} \mathrm{O}(g)\) Balancing O: We have 3 O atoms on the reactant side, which is equal to the number of O atoms on the product side. \(\mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow 2\mathrm{N}_{2} \mathrm{O}(g)+2\mathrm{H}_{2} \mathrm{O}(g)\) Since we have one compound breaking down into simpler compounds, this reaction is a decomposition reaction.
03

Balancing Reaction (c)

Given reaction: \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Balance the elements in the following order: C, H, O. Balancing C: We have 5 C atoms on the reactant side, so we need 5 C atoms on the product side. \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) Balancing H: We have 6 H atoms on the reactant side, so we need 6 H atoms on the product side. \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\) Balancing O: Now we have 10 + 3 = 13 O atoms on the product side, so we need 13/2 O2 molecules on the reactant side. \(\mathrm{C}_{5} \mathrm{H}_{6} \mathrm{O}(l)+\dfrac{13}{2}\mathrm{O}_{2}(g) \longrightarrow 5\mathrm{CO}_{2}(g)+3\mathrm{H}_{2} \mathrm{O}(g)\) Since there is oxygen gas as one of the reactants and the products are water and carbon dioxide, this reaction is a combustion reaction.
04

Balancing Reaction (d)

Given reaction: \(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{NH}_{3}(g)\) Balance the elements in the following order: N, H. Balancing N: We have 2 N atoms on the reactant side, so we need 2 N atoms on the product side. \(\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NH}_{3}(g)\) Balancing H: Now we have 6 H atoms on the product side, so we need 6 H atoms on the reactant side. \(\mathrm{N}_{2}(g)+3\mathrm{H}_{2}(g) \longrightarrow 2\mathrm{NH}_{3}(g)\) As two simple substances are combining to form a more complex product, this reaction is a combination reaction.
05

Balancing Reaction (e)

Given reaction: \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{KOH}(a q)\) Balance the elements in the following order: K, O, H. Balancing K: We have 2 K atoms on the reactant side, so we need 2 K atoms on the product side. \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{KOH}(a q)\) Balancing O: We have 1 O atoms on the reactant side, which is equal to the number of O atoms on the product side. \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{KOH}(a q)\) Balancing H: Now we have 2 H atoms on both sides. \(\mathrm{K}_{2} \mathrm{O}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2\mathrm{KOH}(a q)\) As two simple compounds are combining to form a more complex product, this reaction is a combination reaction.

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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 substance combines with oxygen gas, resulting in the production of energy in the form of heat and light. This is why combustion is often associated with fire. A classic example of a combustion reaction is when hydrocarbons, like in our exercise, react with oxygen to produce carbon dioxide and water.
  • The general formula for hydrocarbon combustion is: \[\text{\text{Hydrocarbon} + O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}\]
  • Key products are always carbon dioxide \( (\text{CO}_2) \) and water \( (\text{H}_2\text{O}) \).
  • These reactions are exothermic, meaning they release energy.
  • In the balanced chemical equations provided, equations (a) and (c) are combustion reactions.
Understanding these chemical processes is important because they illustrate how chemical reactions can transform matter and release energy, providing fuel for engines, heating, and generating electricity.
Decomposition Reactions
Decomposition reactions are a type of chemical reaction where a single compound breaks down into two or more simpler substances. This process is the opposite of a combination reaction. Decomposition reactions can occur through the application of heat, light, or electricity, causing the split of complex molecules.
  • In a general formula, a compound \((\text{AB})\) breaks down into components \((\text{A} + \text{B})\).
  • An example from the exercise is reaction (b), where ammonium nitrate \((\text{NH}_4\text{NO}_3)\) decomposes into nitrous oxide \((\text{N}_2\text{O})\) and water \((\text{H}_2\text{O})\).
  • Such reactions can be spontaneous or require some form of energy to proceed.
  • They play a critical role in different scientific fields, including analysis and waste disposal.
Knowing how these reactions work can help students understand processes like thermal decomposition which is prevalent in both laboratory settings and real-world applications.
Combination Reactions
Combination reactions, also known as synthesis reactions, involve the joining of two or more reactants to form one complex product. These reactions are vital to the formation of more complex molecules from simpler ones, an essential aspect of chemical synthesis.
  • The general formula involves two or more substances \((\text{A} + \text{B})\) combining to form a compound \((\text{AB})\).
  • In our exercise, reactions (d) and (e) are examples of combination reactions:
    • In (d), nitrogen gas \((\text{N}_2)\) and hydrogen gas \((\text{H}_2)\) unite to make ammonia \((\text{NH}_3)\).
    • In (e), potassium oxide \((\text{K}_2\text{O})\) and water \((\text{H}_2\text{O})\) produce potassium hydroxide \((\text{KOH})\).
  • These reactions are fundamental in industrial applications such as the production of fertilizers and chemicals.
  • They often involve energy changes, with energy sometimes being required to drive the reaction forward.
Understanding combination reactions opens students' eyes to the potential of chemical assembly, enabling the creation of new substances critical for technological and industrial advancement.

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Most popular questions from this chapter

One of the steps in the commercial process for converting ammonia to nitric acid is the conversion of \(\mathrm{NH}_{3}\) to \(\mathrm{NO}\) : $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g) $$ In a certain experiment, \(1.50 \mathrm{~g}\) of \(\mathrm{NH}_{3}\) reacts with \(2.75 \mathrm{~g}\) of \(\mathrm{O}_{2}\) (a) Which is the limiting reactant? (b) How many grams of \(\mathrm{NO}\) and of \(\mathrm{H}_{2} \mathrm{O}\) form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conservation of mass.

At least \(25 \mu \mathrm{g}\) of tetrahydrocannabinol (THC), the active ingredient in marijuana, is required to produce intoxication. The molecular formula of \(\mathrm{THC}\) is \(\mathrm{C}_{21} \mathrm{H}_{30} \mathrm{O}_{2}\). How many moles of THC does this \(25 \mu \mathrm{g}\) represent? How many molecules?

(a) Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction almost always less than the theoretical yield? (c) Can a reaction ever have \(110 \%\) actual yield?

A particular coal contains \(2.5 \%\) sulfur by mass. When this coal is burned at a power plant, the sulfur is converted into sulfur dioxide gas, which is a pollutant. To reduce sulfur dioxide emissions, calcium oxide (lime) is used. The sulfur dioxide reacts with calcium oxide to form solid calcium sulfite. (a) Write the balanced chemical equation for the reaction. (b) If the coal is burned in a power plant that uses 2000 tons of coal per day, what mass of calcium oxide is required daily to eliminate the sulfur dioxide? (c) How many grams of calcium sulfite are produced daily by this power plant?

Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a certain number of water molecules are included in the solid structure. Its formula can be written as \(\mathrm{Na}_{2} \mathrm{CO}_{3} \cdot x \mathrm{H}_{2} \mathrm{O}\), where \(x\) is the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). When a \(2.558\) -g sample of washing soda is heated at \(25^{\circ} \mathrm{C}\), all the water of hydration is lost, leaving \(0.948 \mathrm{~g}\) of \(\mathrm{Na}_{2} \mathrm{CO}_{3}\). What is the value of \(x\) ?
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