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### Topic: Balancing really difficult chemical equations  (Read 11753 times)

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#### nectarines

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##### Balancing really difficult chemical equations
« on: December 19, 2012, 08:00:06 PM »
I know this seems extremely simple, but I've been taking practice tests for my gen chem final, and the balancing equations problems have been the most challenging for me. Here are examples:

?P + ?HNO3 + ?H2O -> ?H3PO4 + ?NO

?Cu + ?HNO3 -> ?Cu(NO3)2 + ?NO + ?NO2 + ?H2O, where NO and NO2 are formed in a 2:3 ratio

Can someone explain the process of solving these? I can do most balancing problems, but these are just ridiculous.

#### curiouscat

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##### Re: Balancing really difficult chemical equations
« Reply #1 on: December 20, 2012, 12:10:27 AM »
Have you been taught how to solve simultaneous linear equations?

Then, write atom balances. Say, in 2nd example for Cu, H, N, O.
4 linear eq. in 6 unknowns. Solve.

Matrix based solving might speed up the process.

It isn't the way Chem. classes typically teach but I find it neater. If you're lucky maybe your Scientific Calculator will have a linear eq. solving function too.

#### curiouscat

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##### Re: Balancing really difficult chemical equations
« Reply #2 on: December 20, 2012, 12:35:33 AM »
a Cu + b HNO3    c Cu(NO3)2 + d NO + e NO2 + f H2O

(Cu)  a = c
(H)   b  =  2f
(N)  b = 2c +d + e
(O) 3b = 6c +d + 2e + f

3d=2e

Solve.

#### AWK

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##### Re: Balancing really difficult chemical equations
« Reply #3 on: December 20, 2012, 01:11:36 AM »
Quote
?Cu + ?HNO3 -> ?Cu(NO3)2 + ?NO + ?NO2 + ?H2O, where NO and NO2 are formed in a 2:3 ratio
It will be more simple to split this eqution into two equations with NO and NO2 formed separately.

correction of printing error (quote instead of subscript)
« Last Edit: December 20, 2012, 11:33:16 AM by AWK »
AWK

#### Borek

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##### Re: Balancing really difficult chemical equations
« Reply #4 on: December 20, 2012, 03:26:00 AM »
As AWK wrote, these are two separate processes and each has to be balanced separately.
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#### AWK

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##### Re: Balancing really difficult chemical equations
« Reply #5 on: January 03, 2013, 07:52:00 AM »
In this case the other possibility is joining 2NO + 3NO2 into artificial molecule N5O8
?Cu + ?HNO3 -> ?Cu(NO3)2 + ?N5O8 + ?H2O
AWK

#### Ben Cohen

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##### Re: Balancing really difficult chemical equations
« Reply #6 on: January 10, 2013, 01:22:36 AM »
It's all about making connections between the elements and using variables efficiently

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##### Re: Balancing really difficult chemical equations
« Reply #7 on: January 10, 2013, 12:45:41 PM »
a Cu + b HNO3    c Cu(NO3)2 + d NO + e NO2 + f H2O

(Cu)  a = c
(H)   b  =  2f
(N)  b = 2c +d + e
(O) 3b = 6c +d + 2e + f

3d=2e

Solve.

6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

#### curiouscat

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##### Re: Balancing really difficult chemical equations
« Reply #8 on: January 10, 2013, 01:03:30 PM »
a Cu + b HNO3    c Cu(NO3)2 + d NO + e NO2 + f H2O

(Cu)  a = c
(H)   b  =  2f
(N)  b = 2c +d + e
(O) 3b = 6c +d + 2e + f

3d=2e

Solve.

6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

5 equations. Also have "3d=2e"?

#### Borek

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##### Re: Balancing really difficult chemical equations
« Reply #9 on: January 10, 2013, 01:35:05 PM »
6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

Quite often we do miss one equation when balancing reactions, but we also have additional condition that the solution should consist of the smallest integers - and that's enough to find the unique solution.
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#### curiouscat

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##### Re: Balancing really difficult chemical equations
« Reply #10 on: January 10, 2013, 02:30:50 PM »
6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

This cannot be solved numerically uniquely by the standard method of solving simultaneous linear equations.

An under-determined system of simultanous linear eqns. is not the same as an inconsistent system.

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##### Re: Balancing really difficult chemical equations
« Reply #11 on: January 10, 2013, 03:13:53 PM »
6 variables in 4 equations? This cannot be solved numerically by the standard method of solving simultaneous linear equations.

This cannot be solved numerically uniquely by the standard method of solving simultaneous linear equations.

An under-determined system of simultanous linear eqns. is not the same as an inconsistent system.

What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

#### Borek

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##### Re: Balancing really difficult chemical equations
« Reply #12 on: January 10, 2013, 03:56:08 PM »
What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

I told you - there is an additional condition that can be used. Depending on how you approach the problem you either need some guesswork, or you can do it algorithmically. In a way it is not much different from finding the empirical formula - you have numbers that have to be converted to integers without changing ratios. It is not that difficult, as you can do Gauss elimination using rational numbers, so in the end it is just a matter of finding GCD of all coefficients.

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#### curiouscat

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##### Re: Balancing really difficult chemical equations
« Reply #13 on: January 10, 2013, 05:21:57 PM »
What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

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##### Re: Balancing really difficult chemical equations
« Reply #14 on: January 10, 2013, 05:43:18 PM »
What is the process to force a solution out of something like this?

(When I say force a solution I mean find numerical values by actual number calculation rather than guesswork.)

I told you - there is an additional condition that can be used. Depending on how you approach the problem you either need some guesswork, or you can do it algorithmically. In a way it is not much different from finding the empirical formula - you have numbers that have to be converted to integers without changing ratios. It is not that difficult, as you can do Gauss elimination using rational numbers, so in the end it is just a matter of finding GCD of all coefficients.