Question

How many moles of `CoO(s)` will be formed for every mole of `CoCl_2 * 6H_2O(s)`?

Answer 1

The ratio is `1:1`

Explanation:

Assuming you meant how many moles of `CoO` will be formed from the reaction of `CoCl_2` with a strong base such as `NaOH`, we can start working this out, well, at first we have

`CoCl_2 + NaOH rarr Co(OH)_(2(s)) + NaCl`
(The water was ignored because it's solvent, so, even if we do add it it's not gonna participate in the reaction)

Balancing that equation we see we need to have 2 sodiums in the LHS and 2 chlorides in the RHS, which we can solve by putting a 2 in front of `NaOH` and `NaCl`

`CoCl_2 + 2NaOH rarr Co(OH)_(2(s)) + 2NaCl`

So we know that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) hydroxide, next we need to see the dehydration reaction.

`Co(OH)_2 rarr CoO + H_2O`

Which, if we check is already balanced, so we can say that 1 mole of cobalt (II) chloride makes 1 mole of cobalt (II) oxide.