Diethyl ether, C₄H₁₀O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C₄H₁₀O(l) to CO₂(g) and H₂O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C₄H₁₀O(l).

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Hey everyone today we're being asked to complete and balance this hydrocarbon combustion reaction. Now recall that a combustion reaction occurs when a compound reacts with excess oxygen gas in the air to form products namely Carbon dioxide c. 0 2 And Water H 20. However, specifically in hydrocarbon combustion reactions, we can generally end up with a reaction formula that follows the following pattern. If we have X number of carbons and wine number of hydrogen That react with 02, we can usually or generally will end up with X. c. 0. 2 and why H 20. So utilizing this pattern, we can go ahead and take a look at how our hydrocarbon at our combustion reaction. So in this hydrocarbon we have 12345. So five carbons and 123456789, 10, 10 hydrogen. And this is just simplified. It's not showing us the skeletal structure or anything, but this will help us in the long run and it's reacting with oxygen gas as we mentioned earlier. So, taking a look at the reaction form of that we mentioned earlier. Well, if we have five carbons, then we can logically make five Carbon dioxides from that. Right? So five carbon dioxides and If we have 10 hydrogen as well, H20 has two hydrogen each. So that will generally end up with five Waters, five water molecules. So now that we have our carbon and hydrogen is accounted for, we can take a look at the oxygen's no in C. 02, there's two oxygen's for every molecule of carbon dioxide, which means five carbon dioxides will net us 10 oxygen molecules, sorry. And five waters will notice five more oxygen molecules. Right, that little leader oxygen's. So we'll end up with a total of 15 oxygen's as such. Since we have 02, not just oh alone as a reactant. to end up with 15 oxygen's at the end, We can simply put 15 divided by two. That'll give us 15 singular oxygen's. That'll end up on the product side. However, we generally don't like to have fractions as coefficients in any of these balanced reaction formulas or balanced reactions. Sorry. So to amend this all, we have to do his multiply everything here by two. Multiplied by two. So therefore our balanced reaction will be and I'll write out the whole skeletal structure again. Ch three C H. Wanted to CHCH two. Ch three gas. We'll have two of these plus 15 02 will give us carbon dioxide and 10 water. And this here will be our final balanced reaction. I hope this helps. And I look forward to seeing you in the next one

Diethyl ether, C₄H₁₀O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C₄H₁₀O(l) to CO₂(g) and H₂O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C₄H₁₀O(l).

Verified Solution

Key Concepts

Combustion Reaction

Balancing Chemical Equations

Enthalpy Change (ΔH°)

Diethyl ether, C4H10O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C4H10O(l) to CO2(g) and H2O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C4H10O(l).

Verified Solution

Key Concepts

Combustion Reaction

Balancing Chemical Equations

Enthalpy Change (ΔH°)

Diethyl ether, C₄H₁₀O(l), a flammable compound that was once used as a surgical anesthetic, has the structure The complete combustion of 1 mol of C₄H₁₀O(l) to CO₂(g) and H₂O(l) yields ΔH° = -2723.7 kJ. (a) Write a balanced equation for the combustion of 1 mol of C₄H₁₀O(l).