Answer
a)
$Cu^{2+}$(aq) + $S^{2-}$(aq) -> $CuS$(s)
b)
$Co^{2+}$(aq) + $2OH^{-}$(aq) -> $Co(OH)_{2}$(s)
c)
$Ag^{+}$(aq) + $I^{-}$(aq) -> $AgI$(s)
Work Step by Step
a)
Initially, the reaction is:
$CuSO_{4}$(aq) + $Na_{2}S$(aq) -> $CuS$(s) + $Na_{2}SO_{4}$(aq)
However, since $Na$ and $SO_{4}$ are spectator ions, it has to be removed. Therefore, the reaction turns out to be:
$Cu^{2+}$(aq) + $S^{2-}$(aq) -> $CuS$(s)
b)
Initially, the reaction is:
$CoCl_{2}$(aq) $2NaOH$(aq) -> $Co(OH)_{2}$(s) + $2NaCl$ (aq)
However, since both $Na$ and $Cl$ are spectator ions, the reaction turns out to be:
$Co^{2+}$(aq) + $2OH^{-}$(aq) -> $Co(OH)_{2}$(s)
c)
Initially, the reaction is:
$AgNO_{3}$(aq) + $KI$(aq) -> $AgI$(s) + $KNO_{3}$(aq)
However, since $K^{+}$ and $NO_{3}$$^{-}$ are spectator ions, the reaction turns out to be:
$Ag^{+}$(aq) + $I^{-}$(aq) -> $AgI$(s)