Q4

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tmehrazar
Posts: 39
Joined: Fri Apr 06, 2018 11:04 am

Q4

Postby tmehrazar » Sun Apr 22, 2018 10:09 pm

How do you do question 4 on test 1? Writing a balanced equation for the combustion of fluorene.

Lily Emerson 1I
Posts: 27
Joined: Fri Apr 06, 2018 11:02 am

Re: Q4

Postby Lily Emerson 1I » Sun Apr 22, 2018 10:15 pm

Organic compounds pretty much alway combust into water and co2, so the balances equation is 2C13H10 + 31O2 -> 26CO2 + 10H20

Nahelly Alfaro-2C
Posts: 59
Joined: Wed Nov 15, 2017 3:04 am

Re: Q4

Postby Nahelly Alfaro-2C » Sun Apr 22, 2018 10:19 pm

when you are writing a combustion equation you will always add O2 to the reactant part of the equation because the product will be CO2 and H2O.
So in writing the combustion equation of fluorene, it would be: C13H10+O2----> CO2+H2O. You then have to balance the equation to make it side equal.

Miya Lopez 1I
Posts: 62
Joined: Tue Nov 14, 2017 3:00 am
Been upvoted: 1 time

Re: Q4

Postby Miya Lopez 1I » Sun Apr 22, 2018 10:26 pm

I'm not sure if everyone's tests are in the same order but my Q4 uses Phenanthrene. I wrote a balanced equation for the combustion of Phenanthrene (C14H10).

So first I wrote C14H10 + O2 ---> CO2 + H20

Then I wrote the elements in a column and how much the elements had on each side of the equation, and added coefficients until it was balanced.

14 C 1->14
10 H 2->10
33 <-- 2 O 3-> 7--> 33

I first added a 5 to H20, then a 14 to CO2, then I got 33 O on the right side of the equation and in order to balance it on the left side I put a 33/2 in front of the O2. To make everything whole numbers, I multiplied the whole equation by 2 to get... (coefficients in parentheses)

(2)C14H10 + (33)O2 ---> (28)CO2 + (10)H20


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