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

Mar 8, 2016

The ratio is $1 : 1$

#### Explanation:

Assuming you meant how many moles of $C o O$ will be formed from the reaction of $C o C {l}_{2}$ with a strong base such as $N a O H$, we can start working this out, well, at first we have

$C o C {l}_{2} + N a O H \rightarrow C o {\left(O H\right)}_{2 \left(s\right)} + N a C l$
(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 $N a O H$ and $N a C l$

$C o C {l}_{2} + 2 N a O H \rightarrow C o {\left(O H\right)}_{2 \left(s\right)} + 2 N a C l$

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.

$C o {\left(O H\right)}_{2} \rightarrow C o O + {H}_{2} O$

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.