[jools]

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]

Double check the question: 25.0 g of Al(OH)₃ or Al₄C_3?

[jools]

Double check the question: 25.0 g of Al(OH)₃ or Al₄C_3?

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?

Sorry about that....

[Borek]

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.

[jools]

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?

[Borek]

In fact you need to know only that carbides - when react with water - give hydroxide and some kind of hydrocarbon. If you have one mole of Al₄C₃, how many moles of Al(OH)₃ can be produced? Hint: take a look at number of Al atoms in both substances.