# How do you balance Co(OH)3 + HNO3-->Co(NO3)3+ H2O?

Oct 26, 2015

$C o {\left(O H\right)}_{3} \left(s\right) + 3 H N {O}_{3} \left(a q\right) \rightarrow C o {\left(N O\right)}_{3} \left(a q\right) + 3 {H}_{2} O \left(l\right)$

#### Explanation:

This is effectively an acid base reaction of the sort:

$K O H \left(a q\right) + H N {O}_{3} \left(a q\right) \rightarrow K N {O}_{3} \left(a q\right) + {H}_{2} O \left(l\right)$,

or more simply,

$O {H}^{-} + {H}^{+} \rightarrow {H}_{2} O$

How did I balance it? Well I know that for every species (every atom) on the LHS must correspond to a species on the RHS. And not only must the atoms balance, the electrical charge must balance. Given the 3 equations, are these conditions fulfilled here?

Oct 27, 2015

$C o {\left(O H\right)}_{3} + 3 H N {O}_{3} \to C o {\left(N {O}_{3}\right)}_{3} + 3 {H}_{2} O$

Oct 30, 2015

$C o {\left(O H\right)}_{3} + 3 H N {O}_{3} + C o {\left(N {O}_{3}\right)}_{3} + 3 {H}_{2} O$

#### Explanation:

1 Cobalt on either side so that is fine.
1 Nitrate on the left, 3 on the right so $3 H N {O}_{3}$ on the left.
3 Oxygen on the left (not including the nitrate) , 1 on the right so $3 {H}_{2} O$ on the left
Check hydrogen: 6 on the left and 6 on the right so it's all fine.

Tip: make sure you check that there are an equal number of each element on either side