Balancing equations :(

Question

melissa_something · Answer

$$C_9 H _{20} + O_2 ----> CO_2 +H_2 O$$

melissa_something · Answer

I got the answer, just don't know why my teacher put a 14 in front of O

aaronq · Answer

what do you have so far?

melissa_something · Answer

C_9 H_20 + 26O_2 ----> 9Co_2 + 10H_2O

melissa_something · Answer

@aaronq But instead of the 26 in front of the O my teacher put 14

aaronq · Answer

so on the right side of the equation, you have 9*2+10=28 oxygens

on the left side you have $O_2$ the subscript works like distributive property in math.

$2*(2x)=4x$

same thing in chem

$14~O_2=28~O$

aaronq · Answer

Let me know if that doesn't make sense to you

melissa_something · Answer

Im trying to study sorry, it kinda doesnt make sense

melissa_something · Answer

Where is the 10 on the right side you speak of? @aaronq

aaronq · Answer

$\sf  \large  C_9 H_{20} + 14~O_2 ----> 9~CO_2 + \underbrace{10}_{here}~H_2O$

melissa_something · Answer

Okay I got it! Do you mind explaining the subscript thing again? It would always multiply?

aaronq · Answer

yes, always multiply. Also when a number is not explicitly written as a subscript a 1 is assumed, like in math when you have $x$, it means $x^1$.

so for example, you have $H_2SO_4$

if you have a coefficient (number in front) of 3

$3*(H_2SO_4)$ then you have

3*2=6 H
3*1=3 S
3*4=12 O

it gets a little more complex when you have subscripts in brackets, for example:

$Fe(H_2O)_2Cl$

you have 1 Fe, 4 H's, 2 O'2 and 1 Cl

melissa_something · Answer

Thank you so much :) !!!!!!

aaronq · Answer

no problem (: