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Chapter 5: Problem 101

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

A substance \(\mathrm{X}_{2} \mathrm{Z}\) has the composition (by mass) of \(40.0 \% \mathrm{X}\) and \(60.0 \%\) Z. What is the composition (by mass) of the compound \(\mathrm{XZ}_{2} ?\)

A 2.25-g sample of scandium metal is reacted with excess hydrochloric acid to produce 0.1502 g hydrogen gas. What is the formula of the scandium chloride produced in the reaction?

Aspartame is an artificial sweetener that is 160 times sweeter than sucrose (table sugar) when dissolved in water. It is marketed as NutraSweet. The molecular formula of aspartame is \(\mathrm{C}_{14} \mathrm{H}_{18} \mathrm{N}_{2} \mathrm{O}_{5}.\) a. Calculate the molar mass of aspartame. b. What amount (moles) of molecules are present in \(10.0 \mathrm{g}\) aspartame? c. Calculate the mass in grams of 1.56 mole of aspartame. d. What number of molecules are in \(5.0 \mathrm{mg}\) aspartame? e. What number of atoms of nitrogen are in \(1.2 \mathrm{g}\) aspartame? f. What is the mass in grams of \(1.0 \times 10^{9}\) molecules of aspartame? g. What is the mass in grams of one molecule of aspartame?

A 0.4230-g sample of impure sodium nitrate was heated, converting all the sodium nitrate to 0.2864 g of sodium nitrite and oxygen gas. Determine the percent of sodium nitrate in the original sample.

A gas contains a mixture of \(\mathrm{NH}_{3}(g)\) and \(\mathrm{N}_{2} \mathrm{H}_{4}(g),\) both of which react with \(\mathrm{O}_{2}(g)\) to form \(\mathrm{NO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) The gaseous mixture (with an initial mass of \(61.00 \mathrm{g}\) ) is reacted with 10.00 moles \(\mathrm{O}_{2}\), and after the reaction is complete, 4.062 moles of \(\mathrm{O}_{2}\) remains. Calculate the mass percent of \(\mathrm{N}_{2} \mathrm{H}_{4}(g)\) in the original gaseous mixture.
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