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

Hey everyone. So today we're being asked to complete and balance this hydrocarbon combustion reaction. Now, in general combustion reactions are when a compound reacts with excess oxygen gas in the air to form products namely C. 02 carbon dioxide and water. And for hydrocarbon combustion reactions specifically, we generally have a reaction form that looks somewhat like this. If we have X number of carbons and y number of hydrogen plus oxygen, oxygen gas I should say, would generally end up with XC two plus. Why? H 20? Oh, so with this in mind, we can go ahead and start tackling and balancing this unbalanced reaction that we have right here. So we'll leave the oxygen gas alone for the time being and we'll just take a look at the hydrocarbon. So we have carbons. So let's just write that out real quick. C six and we have 123456789, 10, 10 Hydrogen. Let me write this in blue. So we can differentiate it six carbons and 10 hydrogen. So it's gas Plus 02 gas. Well, if we want to match everything on the right side on the reaction side, and let's just shift this over a little bit. If we want to balance it on the reactant side. Well, if we have six six carbons in the hydrocarbon in the product and that will mean we can yield about six carbon dioxides. and since we have 10 hydrogen in that left hand side hydrocarbon, We can yield about five waters, five molds of water for every one mole of the hydrocarbon. However, with this in mind, now let's go back and take a look at the oxygen because now our carbons and hydrogen are accounted for. But the oxygen's are not. So if we have six C. 02 S and their two oxygen's per carbon dioxide and that means we have 12 oxygen's from here, 12 oxygen's from here and We have five oxygen's from the five waters. So we have a total of 17 oxygen's in the product side. So to balance this out well, If we have 17 oxygen's but 02 is our reactant. Then 17 by two O two's Will give us 17 oxygen's. But this isn't typically how we want our balance equations to look like our balanced reactions. We'd like them to all be in whole numbers so we can go ahead and multiply this entire thing by two times two. And that will give us and I'll just write out the entire hardware carbon again, two ch three CH two. See see ch two ch three gas plus 17 nets 12 carbon dioxides and 10 waters turn H 20. So there you have it. This is our final balanced equation. I hope this helps. And I look forward to seeing you in the next one

Ch.22 - Organic Chemistry

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Ch.22 - Organic Chemistry

Problem 48

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Chapter 22, Problem 48

Complete and balance each hydrocarbon combustion reaction. a. CH₃CH₂CH₂CH₃ + O₂ → b.CH₂=CHCH₃ + O₂ → c. CH≡CCH₂CH₃ + O₂ →

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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. For the given reaction, it would be: CH≡CCH2CH3 + O2 → CO2 + H2O.

Step 3: Balance the carbon atoms first. Count the number of carbon atoms on both sides of the equation and add coefficients to balance them.

Step 4: Next, balance the hydrogen atoms. Again, count the number of hydrogen atoms on both sides of the equation and add coefficients to balance them.

Step 5: Finally, balance the oxygen atoms. This can be a bit tricky because oxygen appears in two compounds on the right side of the equation. You may need to adjust the coefficients you've already set to get the oxygen atoms to balance.

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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, depending on the number of carbon and hydrogen atoms in the hydrocarbon.

Balancing Chemical Equations

Balancing chemical equations is essential to ensure that the number of atoms of each element is the same on both sides of the equation, adhering to the law of conservation of mass. This involves adjusting the coefficients of the reactants and products without changing their chemical identities. For combustion reactions, it often requires balancing carbon, hydrogen, and oxygen atoms systematically.

Hydrocarbon Structure

Hydrocarbons are organic compounds consisting solely of carbon and hydrogen atoms. They can be classified as aliphatic (straight or branched chains) or aromatic (ring structures). The structure of the hydrocarbon affects its combustion reaction; for example, the presence of triple bonds, as in the given compound CH≡CCH2CH3, influences the stoichiometry of the reaction with oxygen.