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Chemistry 13 Online
HanAkoSolo (jamierox4ev3r):

So I'm trying to figure out how to write out the chemical equation for the combustion of isopropyl alcohol

HanAkoSolo (jamierox4ev3r):

I know that the reactants are \(C_3H_8O~+~O_2\) products are: \(H_2O~+~CO_2\)

HanAkoSolo (jamierox4ev3r):

Basically, I'm having trouble balancing this. If someone could give me a quick, foolproof method on balancing any equation in general, that would be fantastic! Also, let me know if my reactants or products are wrong.

HanAkoSolo (jamierox4ev3r):

@abb0t or something...

OpenStudy (abb0t):

First of all, the products for combustion are always: \(\sf \color{red}{CO_2 + H_2O}\) So, you have: \(\sf \color{blue}{C_3H_8O + O_2 \rightarrow CO_2 + H_2O}\) Now, it usually helps if you make a table (or chart), or whatever you want to call it and write down each atom individually and count how many you have on BOTH sides - meaning, reactants and products: \(\sf \color{violet}{reactants \rightarrow products}\) |dw:1392179731969:dw| Notice how there are NOT the same on both sides. To balance, means that both must be equal on both sides! You cannot just input one thing and it just magically disappears. If you make a cookie and add chocolate chips, the chocolate chips dont just disappear on the cookie, they combine to make the cookie!

OpenStudy (abb0t):

So, similar to algebra, you basically play around with the numbers until you get \(\sf \color{red}{SAME}\) number of atoms on BOTH sides!

HanAkoSolo (jamierox4ev3r):

that's a really good cookie analogy @_@ and I understand the concept of balancing and what that entails, sometimes its just doing it that throws me off :P

OpenStudy (abb0t):

It follows the distributive property from algebra, if you put a \(\sf \color{red}{2}\) CO\(_2\), that means now you have 2 C\(_2\) and \(\sf \color{green}{4}\) Oxygen atoms. Because 2 x 1 = 2 and 2 x 2 = 4

HanAkoSolo (jamierox4ev3r):

woah wait a second...*facepalm* I feel like I over thought this...a lot God I'm tired .-.

OpenStudy (abb0t):

Jaime, please. Calm your tits. We can do this, step by step.

HanAkoSolo (jamierox4ev3r):

lol annd you misspelled my name *shrugs* ah whatever, continue on mate :3

OpenStudy (abb0t):

If you multiply the hydrogen by say, 8? I think that might help you...

OpenStudy (abb0t):

Meaning: \(\sf \color{red}{8}\)H\(_2\)O

HanAkoSolo (jamierox4ev3r):

Hydrogen on the right side of the equation? that would automatically mean that Isop. Alcohol would be \(2C_3H_8O\), which also means that on the product side of the equation, you would have \(6O_2\) ...then, from here, hold on let me calculate how much oxygen there would be Sorry for making you wait, the desktop I'm working on keeps freezing

OpenStudy (abb0t):

yES.

HanAkoSolo (jamierox4ev3r):

soo that would mean that there would be \(9O_2\)? ...

OpenStudy (abb0t):

yes.

HanAkoSolo (jamierox4ev3r):

YES success thanks a lot :3 Wow that really wasn't difficult at all, what the heck? LOL oh, and btw my entire family is still pissed off at how badly the broncos lost the superbowl :P

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