confused about this equation 2mno4(-)+16h(+)+10fe(2+)+> 2mn(2+) + 10fe(3+) + 8h20?

Question

first problem is this that in the molecular equation in the right hand side mn(2+) is joined with someother like mnso4 there fore u wrote it alone also what r spectators elements .Also in the molecular equation
kmno4+h2so4=>k2so4+mnso4+3h20 +5(o)
feso4+h2so4+5(o)=> fe2(so4)3+5h2o
main problem is this that in the ionic eqation (the top most) there r 8 h2o and seems that all r from mno4 which produces when 16H joined with o4 of mno4 so how in other molecular eqyations it is 3h2o then 5h2o although when u add it becomes 8h20. the problem is this that first equation shows that all h2o r from mno4 but the two molecular equations show that water is also from mno4 and also from h2so4 thanks

the molecular equation for it is 2kmn4+8h2so4+10fes04=>k2so4+2mnso4+5fe2(so4)3+h2o
which is obtained by adding these two
2kmno4+3h2so4=>k2so4+2mnso4+3h2o+5(O)
10fes04+5h2so4+5(o)=>5fe3(so4)3+5h2o
so here it is shown that first 3hso and in second equation 5 h2o r obtained but in ionic equation it is clear that all the 16 H+ ions join with all the 04 of 2mno4 but in mol.equation 5h2o r obtained by rection of nascent equation with h2so4

Answer 1

I think you would find it easier to be understood if you paid more attention to correct notation, and tried to write clearly.

Your description of the "first problem" doesn't really make sense, but you sound like you're confused about spectator elements and how they are written in the ionic vs. molecular equations.
Molecular equation:
2KMnO4 + 8H2SO4 + 10FeSO4 -> 2MnSO4 + 5Fe2(SO4)3 + K2SO4 + 8H2O
If we break this down into ions we get
2K(+) + 2MnO4(-) + 16H(+) + 8SO4(2-) + 10Fe(2+) + 10SO4(2-) -> 2Mn(2+) + 2SO4(2-) + 10Fe(3+) + 15SO4(2-) + 2K(+) + SO4(2-) + 8 H2O
You can see that on both sides we have 2 K(+) ions and 18 SO4(2-) ions. Since they are unchanged, they are merely spectator ions; if we remove them we get
2MnO4(-) + 16H(+) + 10Fe(2+) -> 2Mn(2+) + 10Fe(3+) + 8H2O
and we can remove a factor of two to get
MnO4(-) + 8H(+) + 5Fe(2+) -> Mn(2+) + 5Fe(3+) + 4H2O
which is the ionic equation.

Now, for the second problem, I'm not sure where you're getting these molecular equations from - they don't balance; balanced versions are
2KMnO4 + 3H2SO4 -> K2SO4 + 2MnSO4 + 3H2O + 5(O)
2FeSO4 + H2SO4 + (O) -> Fe2(SO4)3 + H2O
(or 10FeSO4 + 5H2SO4 + 5(O) -> 5Fe2(SO4)3 + 5H2O to get the same number of loose oxygens)
but to address what seems to be your main point: in the second equation the production of water from the acid requires some loose oxygens, which you can see are coming from the decomposition of the permanganate ion. So ultimately it does all come from the permanganate. What your molecular equations are obscuring is that the iron is needed to supply electrons which are required by the permanganate ion (so that the manganese atom can change from oxidation state 7 to oxidation state 2). You'd get a much better picture of what was going on by writing the redox half-equations, complete with electron flows:

MnO4(-) + 8H(+) + 5e(-) -> Mn(2+) + 4H2O
Fe(2+) -> Fe(3+) + e(-)

to give a balanced redox reaction of
MnO4(-) + 5Fe(2+) + 8H(+) -> Mn(2+) + 5Fe(3+) + 4H2O
as we had before.

Answer 2

Your query is justified.

This ionic equation represents a redox reaction.

Reduction: MnO4- ------> Mn2+
Since, the formal charge of Mn changes from 7 to 2. MnO4- must gain 5 electrons to balance out. In other words, reduction takes place.
MnO4- + 5e- -----> Mn2+

Now, we need to balance the no. of oxygen atoms;
MnO4- + 5e- -----> Mn2+ + 4H20
Now, oxygen is balanced, we need to balance hydrogen.
Since, we're using H2SO4, we'll consider acidic medium(H+).
MnO4- + 5e- + 8H+ -----> Mn2+ + 4H20........(a)


Oxidation: Fe2+ -----> Fe3+
Since, the formal charge changes from +2 to +3, it means in order to balance, Fe3+ must lose 1 electron. In other words, oxidation takes place.
Fe2+------> Fe3+ + e-
Now, we just need to balance the no. of e- in (a). So, we multiply both sides by 5 and we get;
5Fe2+------>5Fe3+ + 5e-............(b)

Now, add (a) and (b). Cancel out that is equal and common just like you do in linear equations.
5Fe2+ + MnO4- + 5e- + 8H+--->Mn2+ + 4H20 + 5Fe3+ + 5e-
5Fe2+ + MnO4- + 8H+-----> Mn2+ + 4H2O + 5Fe3+

This is the actual equation. If, you multiply both sides by 2, you'll get the equation that has put you in doubt.

10Fe2+ + 2MnO4- + 16H+ ---> 2Mn2+ + 8H2O + 10Fe3+

So, you see H+ and MnO4- have given rise to H2O as in equation (a).

Answer 3

haha goodluck

Answer 4

SAME CONDITION HERE ALSO (confused)

Answer 5

Thanks to each and every one of you guys for the replies!

Answer 6

yeah, me too.