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### Topic: Writing equations  (Read 3669 times)

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#### jools

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##### Writing equations
« on: June 22, 2010, 06:19:13 PM »
I have a problem here I was hoping to get a little help with. Here is the question:

What mass of Al(OH)3 can be obtained by reacting 25.0g Al4(OH)3 with water?

In the solution in my textbook it shows the balanced equation as:

Al4C3 + 12H20 --> 3CH4 + 4Al(OH)3

Now... I know how to balance the equation, and I understand the reactants on the left hand side, but I'm a little confused as to how they are getting the methane in the product. Can somebody help me to understand? Thanks!

#### salleebrowne

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##### Re: Writing equations
« Reply #1 on: June 22, 2010, 07:18:31 PM »
Double check the question: 25.0 g of Al(OH)3 or Al4C3?

#### jools

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##### Re: Writing equations
« Reply #2 on: June 22, 2010, 07:26:23 PM »
Double check the question: 25.0 g of Al(OH)3 or Al4C3?

You're absolutely right. The question should have read:

What mass of Al(OH)3 can be obtained by reacting 25.0g Al4C3 with water?

#### Borek

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##### Re: Writing equations
« Reply #3 on: June 23, 2010, 03:18:34 AM »
If you have balanced reaction equation it should be a breeze.

As to why methan is between the products - that's the way some of the carbides (classified as methanides) react with water.
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#### jools

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##### Re: Writing equations
« Reply #4 on: June 23, 2010, 06:49:20 AM »
If you have balanced reaction equation it should be a breeze.

As to why methan is between the products - that's the way some of the carbides (classified as methanides) react with water.

So had I only been given the first half of the problem :

I have a problem here I was hoping to get a little help with. Here is the question:

What mass of Al(OH)3 can be obtained by reacting 25.0g Al4(OH)3 with water?

Is the fact the methane would be produced just something I should already know.... or is there a way to determine what the product will contain?

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