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Balance the following equations representing combustion reactions: c. \(C_{12} H_{22} O_{11}(s)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(g)\) d. \(\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)\) e. \(\operatorname{FeO}(s)+\mathbf{O}_{2}(g) \rightarrow \operatorname{Fe}_{2} \mathbf{O}_{3}(s)\)

Short Answer

Expert verified
The balanced equations for the combustion reactions are as follows: c. \(12C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(g)\) d. \(4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)\) e. \(2FeO(s) + O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\)

Step by step solution

01

c. Balancing combustion of \(C_{12} H_{22} O_{11}\){

}Our initial equation is: \[C_{12} H_{22} O_{11}(s)+O_{2}(g) \rightarrow C O_{2}(g)+H_{2} O(g)\] Step 1: Balance carbon atoms: \[12 C_{12} H_{22} O_{11}(s)+O_{2}(g) \rightarrow 12 CO_{2}(g)+H_{2} O(g)\] Step 2: Balance hydrogen atoms: \[12 C_{12} H_{22} O_{11}(s)+O_{2}(g) \rightarrow 12 CO_{2}(g)+11H_{2}O(g)\] Step 3: Balance oxygen atoms: \[12 C_{12} H_{22} O_{11}(s)+12O_{2}(g) \rightarrow 12 CO_{2}(g)+11H_{2}O(g)\] The balanced equation is: \[12C_{12}H_{22}O_{11}(s) + 12O_{2}(g) \rightarrow 12CO_{2}(g) + 11H_{2}O(g)\]
02

d. Balancing combustion of \(\mathrm{Fe}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{Fe}_{2}\mathrm{O}_{3}\){

}Our initial equation is: \[\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{Fe}_{2}\mathrm{O}_{3}(s)\] Step 1: Balance iron atoms: \[4\mathrm{Fe}(s)+\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\] Step 2: Balance oxygen atoms: \[4\mathrm{Fe}(s)+3\mathrm{O}_{2}(g) \rightarrow 2\mathrm{Fe}_{2}\mathrm{O}_{3}(s)\] The balanced equation is: \[4Fe(s) + 3O_{2}(g) \rightarrow 2Fe_{2}O_{3}(s)\]
03

e. Balancing combustion of \(\mathrm{FeO}\) and \(\mathrm{O}_{2}\) to form \(\mathrm{Fe}_{2}\mathrm{O}_{3}\){

}Our initial equation is: \[\operatorname{FeO}(s)+\mathbf{O}_{2}(g) \rightarrow \operatorname{Fe}_{2}\mathbf{O}_{3}(s)\] Step 1: Balance iron atoms: \[2\operatorname{FeO}(s)+\mathbf{O}_{2}(g) \rightarrow \operatorname{Fe}_{2}\mathbf{O}_{3}(s)\] Step 2: Balance oxygen atoms: \[2\operatorname{FeO}(s)\operatorname{+\mathbf{O}_{2}(g)} \rightarrow \operatorname{Fe}_{2}\mathbf{O}_{3}(s)\] The balanced equation is: \[2FeO(s) + O_{2}(g) \rightarrow Fe_{2}O_{3}(s)\]

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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 chemical processes in which a substance reacts rapidly with oxygen to produce heat and often light. In a combustion reaction, typically a hydrocarbon (containing carbon and hydrogen) or a metal reacts with oxygen to form oxidation products.
As seen in the above exercise, the sugar molecule sucrose \(C_{12}H_{22}O_{11}\), a common carbohydrate, undergoes combustion to form carbon dioxide (\(CO_2\)) and water vapor (\(H_2O\)). The same concept applies to the combustion of iron (\(Fe\)) in the presence of oxygen (\(O_2\)), forming iron(III) oxide (\(Fe_2O_3\)).
Chemical Stoichiometry
Chemical stoichiometry is the component of chemistry that pertains to the calculation of reactants and products in chemical reactions. It allows chemists to determine the quantities of substances consumed and produced in a given reaction based on the law of conservation of mass — matter is neither created nor destroyed.
Within the solution provided, chemical stoichiometric principles are used to determine the amount of oxygen needed to completely react with sucrose and iron, ensuring all atoms are accounted for and the mass remains constant before and after the reaction.
Chemical Reaction Balancing
The process of balancing chemical reactions is integral to chemistry. It involves adjusting the coefficients of the reactants and products to ensure the same number of each type of atom appears on both sides of the equation. This complies with the law of conservation of mass.
For each of the solutions provided, the process starts by balancing the atoms that appear in the least amount of compounds, typically metals or carbon, followed by hydrogens and finally, oxygen atoms. This systematic approach helps organize the balancing process and leads to the correct stoichiometric coefficients.
Oxygen Atoms Balancing
Oxygen atoms balancing is often the final step in balancing combustion reactions due to oxygen's presence in multiple compounds and its diatomic molecular form. After balancing other atoms, the total number of oxygen atoms from all reactants must equal the total number in the products.
The textbook solutions demonstrate this by adjusting the coefficient of oxygen gas after carbon and hydrogen, or iron atoms, are balanced. For instance, with sucrose, once the carbon and hydrogen atoms have been balanced, the final coefficient for oxygen gas is adjusted to ensure that the number of oxygen atoms on both sides of the equation is equal, finalizing the balanced reaction equation.

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Most popular questions from this chapter

Iron oxide ores, commonly a mixture of \(\mathrm{FeO}\) and \(\mathrm{Fe}_{2} \mathrm{O}_{3},\) are given the general formula \(\mathrm{Fe}_{3} \mathrm{O}_{4}\). They yield elemental iron when heated to a very high temperature with either carbon monoxide or elemental hydrogen. Balance the following equations for these processes: $$\begin{array}{c}\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{H}_{2} \mathrm{O}(g) \\\\\mathrm{Fe}_{3} \mathrm{O}_{4}(s)+\mathrm{CO}(g) \longrightarrow \mathrm{Fe}(s)+\mathrm{CO}_{2}(g)\end{array}$$

Give the balanced equation for each of the following. a. The combustion of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) forms carbon dioxide and water vapor. A combustion reaction refers to a reaction of a substance with oxygen gas. b. Aqueous solutions of lead(II) nitrate and sodium phosphate are mixed, resulting in the precipitate formation of lead(II) phosphate with aqueous sodium nitrate as the other product. c. Solid zinc reacts with aqueous HCl to form aqueous zinc chloride and hydrogen gas. d. Aqueous strontium hydroxide reacts with aqueous hydrobromic acid to produce water and aqueous strontium bromide.

The reaction of ethane gas \(\left(\mathrm{C}_{2} \mathrm{H}_{6}\right)\) with chlorine gas produces \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) as its main product (along with HCl). In addition, the reaction invariably produces a variety of other minor products, including \(\mathrm{C}_{2} \mathrm{H}_{4} \mathrm{Cl}_{2}, \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}_{3},\) and others. Naturally, the production of these minor products reduces the yield of the main product. Calculate the percent yield of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}\) if the reaction of \(300 .\) g of ethane with \(650 .\) g of chlorine produced \(490 .\) g of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Cl}.\)

A \(2.077-\mathrm{g}\) sample of an element, which has an atomic mass between 40 and \(55,\) reacts with oxygen to form 3.708 g of an oxide. Determine the formula of the oxide (and identify the element).

Express the composition of each of the following compounds as the mass percents of its elements. a. formaldehyde, \(\mathrm{CH}_{2} \mathrm{O}\) b. glucose, \(C_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) c. acetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\)

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