Complete and balance each hydrocarbon combustion reaction. a. CH 3 CH 2 CH 3 + O 2 → b.CH 3 CH 2 CH=CH 2 +O 2 → c. CH≡CH + O 2 →

All right. Hi, everyone. So the, so this question says that 4 g of elemental sulfur was burned in excess of oxygen. The product was dissolved in enough water to yield a 500 mL solution. Write the balanced chemical equation for this combustion process. So in this case, we happen to have four different answer. Choices. Each with the same atoms or compounds are listed for both the reacting side and the product side. However, what's changing are the stor geometric coefficients in front of my starting materials and my products. Now recall first and foremost that elemental sulfur coming from group six A generally exists as S eight in its most stable form. And because it's elemental sulfur, this is going to be a solid. No, because this is a combustion process, right? Crystalline sulfur or S eight is going to react with oxygen gas to combine together and form sulfur dioxide. Now recall that sulfur dioxide does happen to be a gas with a very characteristic pungent odor making it also a major air pollutant as well. So as of this point, our chemical equation describes the formation of sulfur dioxide by combining elemental sulfur with oxygen gas, however, it's not quite balanced yet, right? Because recall that a chemical equation is only said to be balanced if you have the same number of atoms in the reactant side and the product side. And in this case, we don't happen to have that right. Because if we consider our sulfur, especially, we have eight atoms of sulfur on the left side of the equation and only one atom of sulfur on the right side. So recall that we can fix this by adding a stoichiometry co efficient on the side that has less sulfur atoms. So in this case, I should add a stor chomet coefficient of eight in front of sulfur dioxide on the right side of the equation itself because what that does is actually balance out the number of sulfur atoms that I have on both sides of my equation. However, by doing this, I also unbalance the number of oxygens in this case, right? Because now on the left side, I only have two atoms of oxygen. Whereas on the right side, I happen to have 16 atoms of oxygen because on the right side, I have eight equivalents of sulfur dioxide each with two atoms of oxygen. However, I can fix this in a very similar way that I fix my sulfur or the count of sulfur atoms. Right? Because here I can go ahead and add a stoichiometry coefficient of eight in front of oxygen gas on the left side of the equation. And with that, I've brought, I went ahead and increased the number of oxygen atoms to 16 on the left side, which is equivalent to the number of oxygen atoms on the right side. And there you have it here is your balanced chemical equation. In which one equivalent of solid S eight combines with eight equivalents of oxygen gas to produce eight equivalents of sulfur dioxide, which is also a gas. So the correct answer is going to be option D in the multiple choice. And so with that being said, thank you so very much for watching. And I hope you found this helpful.

Ch.21 - Organic Chemistry

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Tro 4th Edition

Ch.21 - Organic Chemistry

Problem 45

Chapter 21, Problem 45

Complete and balance each hydrocarbon combustion reaction.
a. CH₃CH₂CH₃ + O₂ →
b.CH₃CH₂CH=CH₂ +O₂ →
c. CH≡CH + O₂ →

Verified step by step guidance

Step 1: Identify the type of reaction. In this case, it's a combustion reaction, which typically takes the form of a hydrocarbon reacting with oxygen to produce carbon dioxide (CO2) and water (H2O).

Step 2: Write the unbalanced equation. CH3CH2CH=CH2 + O2 → CO2 + H2O

Step 3: Balance the carbon atoms first. There are 4 carbon atoms on the left side of the equation, so we need 4 CO2 molecules on the right side. CH3CH2CH=CH2 + O2 → 4CO2 + H2O

Step 4: Balance the hydrogen atoms next. There are 8 hydrogen atoms on the left side of the equation, so we need 4 H2O molecules on the right side. CH3CH2CH=CH2 + O2 → 4CO2 + 4H2O

Step 5: Finally, balance the oxygen atoms. There are 16 oxygen atoms on the right side of the equation, so we need 8 O2 molecules on the left side. CH3CH2CH=CH2 + 8O2 → 4CO2 + 4H2O

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Reactions

Combustion reactions involve the reaction of a hydrocarbon with oxygen to produce carbon dioxide and water. These reactions are exothermic, releasing energy in the form of heat and light. In the case of hydrocarbons, the general formula for the products is CO2 and H2O, and balancing these reactions requires ensuring that the number of atoms of each element is equal on both sides of the equation.

Balancing Chemical Equations

Balancing chemical equations is a fundamental skill in chemistry that ensures the law of conservation of mass is upheld. This involves adjusting the coefficients of reactants and products so that the number of atoms for each element is the same on both sides of the equation. For combustion reactions, this often requires balancing carbon, hydrogen, and oxygen atoms systematically.

Alkenes and Their Reactions

Alkenes are hydrocarbons that contain at least one carbon-carbon double bond, which affects their reactivity compared to alkanes. In combustion reactions, alkenes can undergo complete combustion, producing carbon dioxide and water, but they may also participate in other reactions due to the presence of the double bond. Understanding the structure and reactivity of alkenes is crucial for predicting the products of their combustion.