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April 2nd, 2015, 01:08 AM   #1
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How to balance Chemical Equation

Help me to balance the following equation

C3H8 + O2 = H2O + CO2
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April 2nd, 2015, 01:16 AM   #2
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It is very simple Smrithi,

C3H8 + O2 = H2O + CO2

LHS, C = 3, H = 8, O = 2
RHS, C = 1, H = 2, O = 3

Add 3 to CO2,

C3H8 + O2 = H2O + 3CO2
LHS, C = 3, H = 8, O = 2
RHS, C = 3, H = 2, O = 7

Add 4 to H2O,

C3H8 + O2 = 4H2O + 3CO2
LHS, C = 3, H = 8, O = 2
RHS, C = 3, H = 8, O = 10

Add 5 to O2,
C3H8 + 5O2 = 4H2O + 3CO2
LHS, C = 3, H = 8, O = 10
RHS, C = 3, H = 8, O = 10

Hence it is balanced.

Try out Chemical Equation Balancer. It will help you a lot, unless you are well trained.

Last edited by skipjack; July 21st, 2015 at 02:05 AM.
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April 17th, 2015, 07:58 AM   #3
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Typically, when balancing chemical equations you save Oxygen for last. For this equation $C_{3}H_{8} + O_{2} \leftrightharpoons H_{2}O + CO_{2}$ we can solve in 3 steps.
Step 1 balance carbon
$C_{3}H_{8} + O_{2} \leftrightharpoons H_{2}O + 3CO_{2}$
Step 2 balance hydrogen
$C_{3}H_{8} + O_{2} \leftrightharpoons 4H_{2}O + 3CO_{2}$
Step 3 balance oxygen
$C_{3}H_{8} + 5O_{2} \leftrightharpoons 4H_{2}O + 3CO_{2}$
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July 20th, 2015, 11:16 PM   #4
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Math Focus: Calculus
C3H8 + O2 = H2O + CO2
since there are 8 hydrogen on left and 2 on right so as first step you will have to multiply the hydrogen containing compound with 4, since each molecule containing hydrogen contains 2 hydrogens .but doing so will make total 4 oxygen on right, again on left you have 3 carbons, this means so you will have to multiply carbon containing compound with 3.doing this you will get 4+6=10 oxygens, to balance this out just multiply the oxygen containg compond on left with 5, so that it also gets 10 oxygens(each oxygen compound has 2 oxygens).
answer will be
C3H8+5O2=4H2O+3CO2
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November 7th, 2015, 11:31 AM   #5
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Math Focus: Analysis
Quote:
Originally Posted by Smrithi View Post
C3H8 + O2 = H2O + CO2
Quote:
Originally Posted by Sandra View Post
C3H8 + O2 = H2O + CO2
Quote:
Originally Posted by Diehardwalnut View Post
$C_{3}H_{8} + O_{2} \leftrightharpoons H_{2}O + CO_{2}$
Quote:
Originally Posted by manishsqrt View Post
C3H8 + O2 = H2O + CO2
I notice that all of you are using the equals sign on the first step and on every step until the last step. But the equation is only equal on the last step.

The problem with this practice is that people who are not interested in proof but are just skimming your work for an answer might quote you out of context - as I did above - and thus quote you saying something that is not true.

Last edited by Grozny; November 7th, 2015 at 11:59 AM.
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November 7th, 2015, 11:59 AM   #6
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Math Focus: Analysis
Binary mixtures are easy because one can generally solve for one element at a time.

A Monolithic Perspective on Economic Methodology

In this paper I give as my example of the methodology of chemistry – to contrast it with the methodology of economics – a trinary mixture.

Do you want to give it a try, Smrithi?

Find the proportions by weight of

ammonium nitrate, $NH_{4}NO_{3}$

nitromethane, $CH_{3}NO_{2}$

aluminum, $Al$

to produce these products:

water vapor, $H_{2}O$

carbon dioxide, $CO_{2}$

nitrogen, $N_{2}$

aluminum oxide, $Al_{2}O_{3}$
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November 21st, 2015, 04:28 AM   #7
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By considering the oxygen first,

$\text{3NH}_4\text{NO}_3 + \text{2CH}_3\text{NO}_2 \to \text{9H}_2\text{O} + \text{2CO}_2 + \text{4N}_2$

$\text{3NH}_4\text{NO}_3 + \text{2Al} \to \text{6H}_2\text{O} + \text{3N}_2 + \text{Al}_2\text{O}_3$

and as both balance, combine m of the first and n of the second, where (m, n) = 1. Now determine the proportions by weight.
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