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

1 Answer
Mar 8, 2016

Answer:

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.