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Forum Rules: Read This Before Posting Topic: Help with balancing these equations  (Read 9071 times)

0 Members and 1 Guest are viewing this topic. SoulPlaya

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• Mole Snacks: +0/-0 Help with balancing these equations
« on: February 02, 2011, 02:51:57 PM »
Can anyone help me balance this equation? I've gotten the products, but I can't get the stoichiometry.
Al(NO3)3 (aq) + NH3 (aq) + H2O (l) ------> Al(OH)3 (s) + NH4(NO3) (aq)

and

Na[Al(OH)4] (aq) + HNO3 (aq) -----> NaNo3 (aq) + Al(NO3)3 (aq) + H2O (l)

I've been sitting here for a while trying to figure them out, but I've had no luck. Any help would be greatly appreciated. rabolisk

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• Mole Snacks: +45/-25 Re: Help with balancing these equations
« Reply #1 on: February 02, 2011, 04:13:00 PM »
There is no easy way to do these. The first advice I can give you is to balance the elements that only appear in one compound on each side, like aluminum in the first reaction. For the first reaction, it's helpful to think of nitrate (NO3-) as one species rather than breaking it up and balancing nitrogen and oxygen separately.

Try this, and if you are still having problems, I can help you out more. AWK

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« Reply #2 on: February 03, 2011, 03:07:52 AM »
Chemists very often reasoning through analogy.
Could you write down a reaction of Al(NO3)3 (aq) with NaOH? Ammonia in water is also a (weak) base.

Na[Al(OH)4] (aq) is formed from Al(OH)3 and NaOH. Could you write down reaction of both hydroxides with with HNO3?
AWK DevaDevil Re: Help with balancing these equations
« Reply #3 on: February 04, 2011, 02:29:58 PM »
here is my approach when having difficult-tobalance reactions:
make a table with the products and reactants on one side, and the various "building blocks" on the other side:
| Al | NO3 | NH3 | H2O | OH | NH4 |
a Al(NO3)3 | a  | 3a   |        |        |      |        |
b NH3        |    |       |  b     |        |      |        |
c H2O        |    |       |         |   c    |      |        |
d Al(OH)3   |  d |       |        |        |  3d  |       |
e NH4(NO3)|    |   e   |         |        |       |   e  |

Then the only big change in buiding blocks is the following reaction (this you should be able to spot):
H2O + NH3 --> NH4+ + OH-
this means every H2O and each NH3 in the reactants must give one OH- and one NH4+

now, what does this table mean? It gives you a chance to change the reaction to be balanced into an algebraic set of equations (or a matrix, if you are so inclined)

since Al (reactant) = Al (products), a = d
likewize you can deduce:
3a = e
(and since ammonia gives ammonium): b = e
(and since water gives hydroxyl): c = 3d

then you can try to convert the coefficients into each other:
3a = e = b = 3d = c

so if a = 1, then b = 3, c = 3, d = 1 and e = 3

1 Al(NO3)3 (aq) + 3 NH3 (aq) + 3 H2O (l) ------> 1 Al(OH)3 (s) + 3 NH4(NO3) (aq)

sorry for giving the answer on this one, but I hope the way I explained it makes you able to do the other one yourself!

it is a time-consuming way of doing this, but it works for me when the balancing gets really tough Borek Re: Help with balancing these equations
« Reply #4 on: February 04, 2011, 04:09:28 PM »
here is my approach when having difficult-tobalance reactions:

This is so called algebraic method. Try it for this reaction:

MnO4- + H2O2 + H+ -> Mn2+ + O2 + H2O

Note: this is not only a reaction that occurs in reality, it is even used for determination of hydrogen peroxide concentration.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info SoulPlaya

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« Reply #5 on: February 05, 2011, 12:00:34 PM »
here is my approach when having difficult-tobalance reactions:
make a table with the products and reactants on one side, and the various "building blocks" on the other side:
| Al | NO3 | NH3 | H2O | OH | NH4 |
a Al(NO3)3 | a  | 3a   |        |        |      |        |
b NH3        |    |       |  b     |        |      |        |
c H2O        |    |       |         |   c    |      |        |
d Al(OH)3   |  d |       |        |        |  3d  |       |
e NH4(NO3)|    |   e   |         |        |       |   e  |

Then the only big change in buiding blocks is the following reaction (this you should be able to spot):
H2O + NH3 --> NH4+ + OH-
this means every H2O and each NH3 in the reactants must give one OH- and one NH4+

now, what does this table mean? It gives you a chance to change the reaction to be balanced into an algebraic set of equations (or a matrix, if you are so inclined)

since Al (reactant) = Al (products), a = d
likewize you can deduce:
3a = e
(and since ammonia gives ammonium): b = e
(and since water gives hydroxyl): c = 3d

then you can try to convert the coefficients into each other:
3a = e = b = 3d = c

so if a = 1, then b = 3, c = 3, d = 1 and e = 3