Fe(s) + 2CuCl2∙2H2O(aq) = 2Cu(s) + FeCl2∙4H2O(aq)

 

I'm confused with the hydrates......

Are my products correct??

How to balance it???

 

 

Is it: Fe(s) + 2CuCl2∙2H2O(aq) = 2Cu(s) + FeCl2∙4H2O(aq), but the chlorides are not balanced...

How much water is there in the beaker, compared to the amounts in the hydrates?

Fe(s) + 2CuCl2∙2H2O(aq) = 2Cu(s) + FeCl2∙4H2O(aq)

 

the hydrates cannot exist as

 

2CuCl2∙2H2O(aq)

 

if they are dissolved in water. if they are actually aqeous, then the equation becomes:

 

 

Fe(s) + 2CuCl2(aq) = 2Cu(s) + FeCl2(aq)

 

the

 

 

2CuCl2∙2H2O can only exist in a crystalline form not when its dissolved in water.

Fe(s) + 2CuCl2∙2H2O(aq) = 2Cu(s) + FeCl2∙4H2O(aq)

 

the hydrates cannot exist as

 

2CuCl2∙2H2O(aq)

 

if they are dissolved in water. if they are actually aqeous, then the equation becomes:

 

 

Fe(s) + 2CuCl2(aq) = 2Cu(s) + FeCl2(aq)

 

the

 

 

2CuCl2∙2H2O can only exist in a crystalline form not when its dissolved in water.

really??

 

Hmmmmmm....... but my teacher said that we need to change the cupric chloride into a hydrate, and after we change it, we will predict the products.

Also she mentioned that the iron chloride will become hydrate if the reaction occurs. That's why I'm confused.

 

How much water is there in the beaker, compared to the amounts in the hydrates?

we did not any water, its already in aqueous solution when my teacher gave it to us..

"we did not any water, its already in aqueous solution when my teacher gave it to us.."

It's already in solution.

So the answer to my question was

"there's a lot of water."

So trying to account for the tiny amount of water present as water of hydration is pointless.

 

Once it's in solution the water in the "2CuCl2∙2H2O(aq)" is meaningless- not least because all the water gets mixed up.

 

What's actually present in the solution would be better described as [Cu(H2O)n]++

where n is probably about 5 or 6

 

Hydrates only really make sense as solids.

Yes, you are right.

Hmmmmmm.... maybe I'll ask my teacher about this.

Thanks for the help:)