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Reaction Aluminium Selenide & Water


Guest adeyave

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Guest adeyave

As part of a science project I need to produce a balanced equation for the reaction between Aluminium Selenide and Water producing the solid Aluminium Hydroxide and Hydrodgen Selenide.

My calculation is in the following format:

 

Al2 SE3 + H20 = H2SE + AL(OH)3

 

Add the number 2 b4 al on the right to balance the aluminium

AL 2 SE3 + H20 = H2 SE3 + 2AL(OH)3

 

Then add the number three b4 h2 se3 to balance the selenium

 

AL2 SE3 + H20 = 3H2 SE3 + 2AL(OH)3

 

To balance the hydrodgen add the number six b4 h20 on the left

 

AL2 SE3 + 6H20 = 3H2 SE + 2AL(OH)3

 

This then leaves us with six oxygen atoms on both sides and hopefully a balanced equation.

 

Am I right?

 

An early reply would be appreciated I am working to a tight deadline

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It looks correct to me. By the way, on the last line, which is the correct balances equation, you missed out a 3 after the SE. You said:

 

AL2 SE3 + 6H20 = 3H2 SE + 2AL(OH)3

It should actaully be:

AL2 SE3 + 6H20 = 3H2 SE3 + 2AL(OH)3

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Al2Se3 + 6H2O --›3H2Se + 2Al(OH)3 Is the correct answer.

 

Aom: It should actaully be:

AL2 SE3 + 6H20 = 3H2 SE3 + 2AL(OH)3

 

That would give you 6 molecules of a 3 molar compund, you need 3 moles of a single.

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