Complete and balance each hydrocarbon combustion reaction.
c. CH‚...

Question

Complete and balance each hydrocarbon combustion reaction.
c. CH‚CCH2CH3 + O2 ¡

Accepted Answer

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

Complete and balance each hydrocarbon combustion reaction. c. CH‚CCH2CH3 + O2 ¡

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