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Balancing Chemical Equations


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Catherine43
Subject: Balancing Chemical EquationsQuote this post in your reply
I dunno if it is a right place to post this, but hope will get help from this forum members.

I need to balance a chemical equation

CH3OH + O2 = CO2 + H2O

I don want any websites for the answer, as i already got it from https://www.easycalculation.com/chem...-equations.php

Kindly some one well known , can say the steps.

It is my asignment problem

Thanks in advance
Posted: 2015-07-17 08:18:32
Ursus
Subject: Re: Balancing Chemical EquationsQuote this post in your reply
Hello Catherine, I’m not a native English-speaker, so my apologies if I make grammatical mistakes, but i'll try to explain a quick method to chemical balancing.

You have to balance the chemical reaction of: CH3OH + O2 => CO2 + H2O
(don't use the = mark, but make it an arrow please smile )

The number 3 in CH3OH is called an 'index-number' and indicates that there are 3 H-atoms on that spot. These numbers may NEVER be changed because if you do, you change the Chemicals. The only numbers that you may change are the numbers in front of the chemical formula’s.

The total of atoms in that molecule equals:
(1 x C) + (3 x H) + (1 x O ) + (1 x H) = 6 atoms.

When you add a molecule O2, then you have 8 atoms when you start the reaction and after the reaction you have to get all the atoms back (but in another composition).

Step 1:
CH3OH + O2 => CO2 + H2O
Before reaction => after reaction
C= 1 C= 1
O= 1 + 2 O= 2 + 1
H= 4 H= 2 <== problem… 4 times an H atom before, only 2 after the reaction.

Because of the fixed index-numbers the only way to even out these atoms is to double amount of the H2O molecule.


Step 2:
CH3OH + O2 => CO2 + 2 H2O
Before reaction => after reaction
C= 1 C= 1
H= 4 H= 4 <== problem fixed.

O= 1 + 2 O= 2 + 2 <== NEW problem… 3 times an O atom before, 4 needed after the reaction.


We need to add one oxygen-atom before the reaction. Oxygen-molecules are 2 oxygen-atoms, so that’s not a simple option. What we need is to add half a molecule.


Step 3:
CH3OH + 1½ O2 => CO2 + 2 H2O
Before reaction => after reaction
C= 1 C= 1
H= 4 H= 4 <== problem fixed.

O= 1 + 3 O= 2 + 2 <== problem fixed.


There is only one last problem: the numbers before the chemical-notations can never be anything but whole numbers. So in order to fix that problem, you double everything…

Step 3:
Before: CH3OH + 1½ O2 => CO2 + 2 H2O => after: 2 CH3OH + 3 O2 => 2 CO2 + 4 H2O


I hope you get it and found this helpful. wink
Posted: 2016-02-25 06:26:04
 
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