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Chapter 3: Problem 21

Balance the following equations and indicate whether they are combination, decomposition, or combustion reactions: (a) \(\mathrm{C}_{7} \mathrm{H}_{16}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\) (b) \(\mathrm{Li}_{3} \mathrm{~N}(s)+\mathrm{BN}(s) \longrightarrow \mathrm{Li}_{3} \mathrm{BN}_{2}(s)\) (c) \(\mathrm{Zn}(\mathrm{OH})_{2}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\) (d) \(\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Ag}(s)+\mathrm{O}_{2}(g)\)

Short Answer

Expert verified
(a) Combustion, balanced: \(\mathrm{C}_{7}\mathrm{H}_{16} + 11\mathrm{O}_{2} \rightarrow 7\mathrm{CO}_{2} + 8\mathrm{H}_{2}\mathrm{O}\). (b) Combination, balanced as is. (c) Decomposition, balanced as is. (d) Decomposition, balanced: \(2\mathrm{Ag}_{2}\mathrm{O} \rightarrow 4\mathrm{Ag} + \mathrm{O}_{2}\).

Step by step solution

01

Analyze Reaction (a)

Equation (a) is a combustion reaction because it involves a hydrocarbon reacting with oxygen to form carbon dioxide and water. Combustion reactions typically have the form of a hydrocarbon (or similar) plus oxygen producing CO鈧� and H鈧侽.
02

Balance Equation (a)

The unbalanced equation is \(\mathrm{C}_{7} \mathrm{H}_{16}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(l)\). First, balance carbon atoms: 7 carbon atoms from \(\mathrm{C}_7\mathrm{H}_{16}\) gives 7 \(\mathrm{CO}_2\). Next, balance hydrogen: 16 hydrogens give 8 \(\mathrm{H}_2\mathrm{O}\). Finally, balance oxygen with 11 \(\mathrm{O}_2\): 7 CO鈧� needs 14 O and 8 H鈧侽 needs 8 O, totaling 22 O in products, equivalent to 11 O鈧�. The balanced equation is: \(\mathrm{C}_{7}\mathrm{H}_{16}(s) + 11\mathrm{O}_{2}(g) \longrightarrow 7\mathrm{CO}_{2}(g) + 8\mathrm{H}_{2}\mathrm{O}(l)\).
03

Analyze Reaction (b)

Equation (b) is a combination reaction because two reactants, \(\mathrm{Li}_{3} \mathrm{~N}\) and \(\mathrm{BN}\), are combining to form a single product, \(\mathrm{Li}_{3} \mathrm{BN}_{2}\).
04

Balance Equation (b)

The unbalanced equation is \(\mathrm{Li}_{3} \mathrm{~N}(s)+\mathrm{BN}(s) \longrightarrow \mathrm{Li}_{3} \mathrm{BN}_{2}(s)\). Balance the equation by ensuring each type of atom is present in equal quantities on both sides: 1 \(\mathrm{Li}_{3} \mathrm{~N} + 1\mathrm{BN} \rightarrow 1 \mathrm{Li}_{3} \mathrm{BN}_{2}\). The equation is already balanced as written.
05

Analyze Reaction (c)

Equation (c) is a decomposition reaction because a single compound, \(\mathrm{Zn(OH)}_2\), breaks down into two products, \(\mathrm{ZnO}\) and \(\mathrm{H}_2\mathrm{O}\).
06

Balance Equation (c)

The unbalanced equation is \(\mathrm{Zn(OH)}_2(s) \longrightarrow \mathrm{ZnO}(s) + \mathrm{H}_2\mathrm{O}(l)\). To balance, note that each side already contains the same numbers of each type of atom: 1 Zn, 2 H, and 2 O from \(\mathrm{Zn(OH)}_{2}\) match those in \(\mathrm{ZnO}\) and \(\mathrm{H}_2\mathrm{O}\). The equation is already balanced.
07

Analyze Reaction (d)

Equation (d) is a decomposition reaction because \(\mathrm{Ag}_2\mathrm{O}\) is breaking down into two products, \(\mathrm{Ag}\) and \(\mathrm{O}_2\).
08

Balance Equation (d)

The unbalanced equation is \(\mathrm{Ag}_2\mathrm{O}(s) \longrightarrow \mathrm{Ag}(s) + \mathrm{O}_2(g)\). To balance, note there are 2 silver atoms in \(\mathrm{Ag}_2\). Add 2 \(\mathrm{Ag}\) to products: \(2\mathrm{Ag}_2\mathrm{O} \longrightarrow 4\mathrm{Ag} + \mathrm{O}_2\). This balances both sides: 4 Ag and 2 O in reactants and products.

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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 a common type of chemical reaction where a substance reacts with oxygen to release energy in the form of heat or light. These reactions usually involve hydrocarbons鈥攃ompounds made of hydrogen and carbon鈥攁nd result in the formation of carbon dioxide and water. A typical example can be seen in reaction (a):
\[\mathrm{C}_{7} \mathrm{H}_{16}(s)+11 \mathrm{O}_{2}(g) \longrightarrow 7 \mathrm{CO}_{2}(g)+8 \mathrm{H}_{2} \mathrm{O}(l)\]
Here's what happens in steps:
  • Identify the Reactants and Products: In combustion reactions, identify the hydrocarbon and oxygen as reactants, and carbon dioxide and water as products.
  • Balance the Carbon: Adjust coefficients to balance the number of carbon atoms on both sides of the equation.
  • Balance the Hydrogen: Then balance the hydrogen atoms, typically by adjusting the water molecules produced.
  • Balance the Oxygen: Finally, count the oxygen molecules required from the products and adjust the \( \mathrm{O}_2 \) molecules in the reactants accordingly.
These stepwise balances ensure that the mass and atom count are equal on both sides of the equation, sticking to the law of conservation of mass.
Combination Reactions
In combination reactions, also known as synthesis reactions, two or more simple substances combine to form a more complex compound. This is exemplified in reaction (b):
\[\mathrm{Li}_{3} \mathrm{~N}(s)+\mathrm{BN}(s) \longrightarrow \mathrm{Li}_{3} \mathrm{BN}_{2}(s)\]
Here's a breakdown of the steps involved:
  • Identify the Reactants and Product: Identify the reactants that merge to form a single compound.
  • Balancing the Equation: Since synthesis reactions usually become balanced naturally as they form one simple compound, you often find the equation balanced with equal counts of each atom on both reactants and product sides from the start.
Understanding these reactions is crucial, as they appear in many practical applications, such as the formation of compounds like salts or combining elements to form compounds with desired properties.
Decomposition Reactions
Decomposition reactions involve breaking down one compound into two or more simpler substances. Reaction (c) and (d) illustrate this process:
(c) \(\mathrm{Zn(OH)}_{2}(s) \longrightarrow \mathrm{ZnO}(s)+\mathrm{H}_{2} \mathrm{O}(l)\)
(d) \(\mathrm{Ag}_{2} \mathrm{O}(s) \longrightarrow \mathrm{Ag}(s)+\mathrm{O}_{2}(g)\)
The process of decomposition usually involves a single reactant yielding multiple products. Steps to balance include:
  • Identify the Breakdown: Recognize when a compound splits into simpler components.
  • Balancing the Products: Ensure the number of atoms for each element in the reactants matches those in the products. Adjust coefficients as needed.
  • Special Considerations: Sometimes, decomposition requires energy like heat, light, or electricity to proceed.
This type of reaction is quite common, especially in laboratory settings and industrial processes where compounds need to be broken down into more useful forms.

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Most popular questions from this chapter

Determine the empirical formulas of the compounds with the following compositions by mass: (a) \(74.0 \% \mathrm{C}, 8.7 \% \mathrm{H},\) and \(17.3 \% \mathrm{~N}\) (b) \(57.5 \% \mathrm{Na}, 40.0 \% \mathrm{O},\) and \(2.5 \% \mathrm{H}\) (c) \(41.1 \% \mathrm{~N}, 11.8 \% \mathrm{H},\) and the remainder \(\mathrm{S}\)

Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction: $$ 8 \mathrm{H}_{2} \mathrm{~S}(g)+4 \mathrm{O}_{2}(g) \longrightarrow \mathrm{S}_{8}(l)+8 \mathrm{H}_{2} \mathrm{O}(g) $$ Under optimal conditions the Claus process gives \(98 \%\) yield of \(S_{8}\) from \(\mathrm{H}_{2} \mathrm{~S}\). If you started with \(30.0 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{~S}\) and \(50.0 \mathrm{~g}\) of \(\mathrm{O}_{2},\) how many grams of \(S_{8}\) would be produced, assuming \(98 \%\) yield?

Calcium hydride reacts with water to form calcium hydroxide and hydrogen gas. (a) Write a balanced chemical equation for the reaction. (b) How many grams of calcium hydride are needed to form \(4.500 \mathrm{~g}\) of hydrogen?

Write a balanced chemical equation for the reaction that occurs when (a) titanium metal reacts with \(\mathrm{O}_{2}(g) ;(\mathbf{b})\) silver(I) oxide decomposes into silver metal and oxygen gas when heated; \((\mathbf{c})\) propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}(l)\) burns in air;(d) methyl tert-butyl ether, \(\mathrm{C}_{5} \mathrm{H}_{12} \mathrm{O}(l),\) burns in air.

When hydrocarbons are burned in a limited amount of air, both \(\mathrm{CO}\) and \(\mathrm{CO}_{2}\) form. When \(0.450 \mathrm{~g}\) of a particular hydrocarbon was burned in air, \(0.467 \mathrm{~g}\) of \(\mathrm{CO}, 0.733 \mathrm{~g}\) of \(\mathrm{CO}_{2}\), and \(0.450 \mathrm{~g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of \(\mathrm{O}_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?
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