Question #929f6

The $C {o}^{2 +} \mathmr{and} P {O}_{4}^{3 -}$ ions will form a precipitate.
So we can leave out the $C {l}^{-} \mathmr{and} N {a}^{+}$ ions, as they are only 'spectators', i.e. they will be left unchanged in the solution.
Since $C o$ has a charge of +2 and $P {O}_{4}$ has a charge of -3, we will have to take 3 of the first and 2 of the other to make charges equal (but opposite):
$3 C {o}^{2 +} \left(a q\right) + 2 P {O}_{4}^{3 -} \left(a q\right) \to C {o}_{3} {\left(P {O}_{4}\right)}_{2} \left(s\right)$