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If \(2.0 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{COOH}, 2.0 \mathrm{~mol} \mathrm{C}_{4} \mathrm{H}_{10},\) and \(2.0 \mathrm{~mol}\) \(\mathrm{C}_{6} \mathrm{H}_{6}\) are completely combusted in oxygen, which one produces the largest number of moles of \(\mathrm{H}_{2} \mathrm{O}\) ? Which one produces the least? Explain.

Short Answer

Expert verified
The compound that produces the largest number of moles of H鈧侽 upon complete combustion is C鈧凥鈧佲個 (butane), with 10.0 moles of H鈧侽, while the compound that produces the least number of moles of H鈧侽 is C鈧咹鈧 (benzene), with 6.0 moles of H鈧侽.

Step by step solution

01

Write balanced chemical equations for the combustion reactions

For each compound, we can write a general equation for its combustion reaction: Compound + O鈧 鈫 CO鈧 + H鈧侽 Then, we balance the equations: 1. For the combustion of CH鈧僀H鈧侰H鈧侰OOH (butanoic acid): \( C_4H_8O_2 + O_2 \rightarrow CO_2 + H_2O \) Balancing this equation, we get: \( C_4H_8O_2 + 6O_2 \rightarrow 4CO_2 + 4H_2O \) 2. For the combustion of C鈧凥鈧佲個 (butane): \( C_4H_{10} + O_2 \rightarrow CO_2 + H_2O \) Balancing this equation, we get: \( 2C_4H_{10} + 13O_2 \rightarrow 8CO_2 + 10H_2O \) 3. For the combustion of C鈧咹鈧 (benzene): \( C_6H_6 + O_鈧 \rightarrow CO鈧 + H鈧侽 \) Balancing this equation, we get: \( 2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O \)
02

Determine moles of H鈧侽 produced by each compound

Now, for each combustion reaction equation, we calculate the number of moles of H鈧侽 produced: 1. For the combustion of 2.0 moles of CH鈧僀H鈧侰H鈧侰OOH: \( 2.0\,\text{mol}\,C_4H_8O_2 \times \frac{4\,\text{mol}\,H_2O}{1\,\text{mol}\,C_4H_8O_2} = 8.0\,\text{mol}\,H_2O \) 2. For the combustion of 2.0 moles of C鈧凥鈧佲個: \( 2.0\,\text{mol}\,C_4H_{10} \times \frac{10\,\text{mol}\,H_2O}{2\,\text{mol}\,C_4H_{10}} = 10.0\,\text{mol}\,H_2O \) 3. For the combustion of 2.0 moles of C鈧咹鈧: \( 2.0\,\text{mol}\,C_6H_6 \times \frac{6\,\text{mol}\,H_2O}{2\,\text{mol}\,C_6H_6} = 6.0\,\text{mol}\,H_2O \)
03

Compare moles of H鈧侽 to find the largest and smallest producers

The number of moles of H鈧侽 produced in each case are: - For CH鈧僀H鈧侰H鈧侰OOH: 8.0 moles of H鈧侽 - For C鈧凥鈧佲個: 10.0 moles of H鈧侽 - For C鈧咹鈧: 6.0 moles of H鈧侽 Therefore, the compound that produces the largest number of moles of H鈧侽 is C鈧凥鈧佲個 (butane), and the compound that produces the least number of moles of H鈧侽 is C鈧咹鈧 (benzene).

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Key Concepts

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

Understanding Stoichiometry
Stoichiometry is the branch of chemistry that involves calculating the quantities of reactants and products in chemical reactions. It's like a recipe for chemical equations, ensuring that everything balances. Using stoichiometry, we determine how different elements and compounds will interact and what they will produce.

Consider a combustion reaction, where a compound reacts with oxygen to produce carbon dioxide and water. To calculate the amount of water produced, we rely on stoichiometry. This involves using the coefficients from a balanced chemical equation to relate the quantities of one substance to another.

Let's break it down:
  • Identify the balanced chemical equation. For example, the complete combustion of butane requires this equation: \( 2C_4H_{10} + 13O_2 ightarrow 8CO_2 + 10H_2O \).
  • Use mole ratios to connect the number of moles of reactants to the number of moles of products. The coefficients in the equation tell us that 2 moles of butane produce 10 moles of water.


With this understanding, you can calculate exactly how much of each product is formed from a given amount of reactant.
Deciphering Chemical Equations
Chemical equations are symbolic representations of chemical reactions where the substances involved are expressed with their chemical formulas. These equations are vital for visualizing the transformation of molecules in a reaction.

Here's what you need to know to decipher a chemical equation:
  • The left side contains the reactants, and the right side lists the products.
  • The arrow signifies the direction of the reaction.
  • Coefficients before each formula show the molar quantities needed.
Every atom on the reactant side must appear equally on the product side, a principle known as "conservation of mass." This leads us to balance equations carefully.

For example, in butanoic acid's combustion, the unbalanced equation simply looks like: \( C_4H_8O_2 + O_2 ightarrow CO_2 + H_2O \).

Through balancing, it becomes: \( C_4H_8O_2 + 6O_2 ightarrow 4CO_2 + 4H_2O \). This balanced equation ensures that all atoms from reactants reappear intact in the products. Such equations allow chemists to accurately predict the outcomes of reactions, making chemical equations fundamental in chemistry.
Mastering the Mole Concept
The mole is a fundamental concept in chemistry, serving as a bridge between the atomic and macroscopic worlds. It's a unit of measurement that allows chemists to count particles, such as atoms, molecules, or ions, in a given substance.

One mole is defined as exactly \( 6.022 \times 10^{23} \) of these entities. This immense number, known as Avogadro's number, enables chemists to perform calculations that involve enormous quantities of atoms and molecules.

In terms of combustion, like our example with benzene \((C_6H_6)\), understanding moles helps determine product amounts. For instance, in the reaction \( 2C_6H_6 + 15O_2 \rightarrow 12CO_2 + 6H_2O \), moles clarify conversions. If you start with 2.0 moles of benzene, you can use its mole ratio with water (from the balanced equation) to calculate that 6.0 moles of water will result.

Grasping the mole concept is essential for conversions between mass, number of particles, and volume, making it a cornerstone of stoichiometry and chemical reactions.

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Most popular questions from this chapter

Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately \(300 \mathrm{mg}\) HCN per kilogram of air when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measuring \(3.5 \times 4.5 \times 2.5 \mathrm{~m}\). The density of air at \(26^{\circ} \mathrm{C}\) is \(0.00118 \mathrm{~g} / \mathrm{cm}^{3} .(\mathbf{b})\) If the HCN is formed by reaction of \(\mathrm{NaCN}\) with an acid such as \(\mathrm{H}_{2} \mathrm{SO}_{4},\) what mass of NaCN gives the lethal dose in the room? $$ 2 \mathrm{NaCN}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+2 \mathrm{HCN}(g) $$ (c) HCN forms when synthetic fibers containing Orlon \(^{\text {- }}\) or Acrilan \(^{\circledast}\) burn. Acrilan \(^{\circledast}\) has an empirical formula of \(\mathrm{CH}_{2} \mathrm{CHCN},\) so HCN is \(50.9 \%\) of the formula by mass. A rug measures \(3.5 \times 4.5 \mathrm{~m}\) and contains \(850 \mathrm{~g}\) of Acrilan \(^{\circledast}\) fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is \(20 \%\) and that the carpet is \(50 \%\) consumed.

What is the mass, in kilograms, of an Avogadro's number of people, if the average mass of a person is \(160 \mathrm{lb}\) ? How does this compare with the mass of Earth, \(5.98 \times 10^{24} \mathrm{~kg}\) ?

Calculate the following quantities: (a) mass, in grams, of \(1.50 \times 10^{-2} \mathrm{~mol} \mathrm{CdS}\) (b) number of moles of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(86.6 \mathrm{~g}\) of this substance (c) number of molecules in \(8.447 \times 10^{-2} \mathrm{~mol} \mathrm{C}_{6} \mathrm{H}_{6}\) (d) number of \(\mathrm{O}\) atoms in \(6.25 \times 10^{-3} \mathrm{~mol} \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\)

(a) One molecule of the antibiotic penicillin G has a mass of \(5.342 \times 10^{-21} \mathrm{~g}\). What is the molar mass of penicillin G? (b) Hemoglobin, the oxygen-carrying protein in red blood cells, has four iron atoms per molecule and contains \(0.340 \%\) iron by mass. Calculate the molar mass of hemoglobin.

Viridicatumtoxin B, \(\mathrm{C}_{30} \mathrm{H}_{31} \mathrm{NO}_{10},\) is a natural antibiotic compound. It requires a synthesis of 12 steps in the laboratory. Assuming all steps have equivalent yields of \(85 \%\), which is the final percent yield of the total synthesis?

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