# Is a hydrate affected by a coeffiecient?

## Homework Statement

Ba(OH)2 · 8 H2O(s) + NH4SCN(s) → Ba(SCN)2(s) + H2O(l) + NH3(g)

## Homework Equations

i have to balance this reaction

## The Attempt at a Solution

2Ba(OH)2 * 8H2O(s) + 4NH4SCN(s) --> 2Ba(SCN)2(s) +12H2O(l) + 4NH3(g)

this isnt the right answer cuz the webassign keeps marking it wrong. i think it boils down to whether a coefficient affects a hydrate or not but idk. thx in advance for any tips u can provide :)

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Notice that all the coefficients in your answer are divisible by 2.

reducing it to 0:2:6:2 isnt the answer

Last edited:
Of course it's not. 2 divided by 2 does not equal 0.

symbolipoint
Homework Helper
Gold Member
Your given set of reactants and products shows bound water on the left side, but unbounded water on the products side. This will help you know how many of the waters to tell on the right side (products).

You seem to recognize that the Ba+2 needs two SCN-, so what should this mean for the count of NH4(SCN) on the left side?

DDT: i meant 0 as in no coefficient. 1:2:6:2 is not the answer, regardless of my imprecise notation

Thanks for the help, Symbollipoint.

chemisttree
Homework Helper
Gold Member

## Homework Statement

Ba(OH)2 · 8 H2O(s) + NH4SCN(s) → Ba(SCN)2(s) + H2O(l) + NH3(g)

remember that Ba(OH)2 · 8 H2O(s) is Ba+2 + 2OH- + 8H2O

and 2 [Ba(OH)2 · 8 H2O(s)] is 2Ba+2 + 4OH- + 16H2O

There is always a 1:8 ratio of Ba to H2O in this hydrate. You don't separate them. You put a coefficient in front of the Barium and it affects the number of water molecules as well.