How to do 16th question?

You'll have to write out the half-reactions to figure this out... except we'll write the proper phases...

#2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq)#

REDUCTION HALF-REACTION

In either basic or acidic media, permanganate gets reduced to manganese (IV) oxide, and then #"Mn"^(2+)# if it goes all the way:

#stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s)#

This is of course containing NORMAL #""^(16) "O"#... Here we choose basic media... balance the oxygen.

#stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)#

Balance hydrogens...

#4stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)#

Balance charge...

#3e^(-) + 4stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l)#

That's the reduction half-reaction, which must be added to an oxidation half-reaction.

OXIDATION HALF-REACTION

Hence, we proceed to the oxidation of #""^(1)"H"_2""^(18)"O"_2# to see where those oxygen atoms go...

#stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g)#

Here we KNOW it must become #""^(18)"O"_2# and NOT #""^(18)"O"""^(1)"H"^(-)#, as we KNOW that oxidation occurs here.

Balance the hydrogens...

#stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(+1))(""^(1)"H"^(+))(aq)#

Balance the charge...

#stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(+1))(""^(1)"H"^(+))(aq) + 2e^(-)#

OVERALL REACTION

And now we add them.

#2(cancel(e^(-)) + 4stackrel(color(blue)(+1))("H"^(+))(aq) + stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) -> stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l))#
#ul(3(stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(+1))(""^(1)"H"^(+))(aq) + cancel(2e^(-))))#
#2stackrel(color(blue)(+1))("H"^(+))(aq) + 2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 4stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g)#

And of course, we're in basic media, so we add the usual #"OH"^(-)# to both sides...

#cancel(2stackrel(color(blue)(+1))("H"^(+))(aq) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq))^(2"H"_2"O"(l)) + 2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + cancel(4)^(2)stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq)#

And our final result is then...

#bb(2stackrel(color(blue)(+7))("Mn")stackrel(color(blue)(-2))("O"_4^(-))(aq) + 3stackrel(color(blue)(+1))(""^(1)"H"_2)""^(18)stackrel(color(blue)(-1))("O"_2)(aq) -> 2stackrel(color(blue)(+4))("Mn")stackrel(color(blue)(-2))("O"_2)(s) + 2stackrel(color(blue)(+1))("H"_2)stackrel(color(blue)(-2))("O")(l) + 3stackrel(color(blue)(0))(""^(18)"O"_2)(g) + 2stackrel(color(blue)(-2))("O")stackrel(color(blue)(+1))("H"^(-))(aq))#