Balance the following equation. (for a balanced eq. aA + bB → cC + dD, enter your answer as the integer abcd)
MnO4−(aq) + SO32−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
I have tried putting in diff answers but its not working
This is a redox, you have to balance charge. Have you studied redox?
origSpecies NewSpecies change
Mn+7.............Mn+2 .......gained 5 electrons
S+4...............S+6.......lost two electrons
SO,you will need to Mn for each five S
2MnO4−(aq) + 2.5(SO3)2−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
is the first go. Then start balancing mass then get rid of the fraction..
4MnO4−(aq) + 5(SO3)2−(aq) + H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + H2O(l)
then balance the O (dont mess with any S related O, so balance the water O
4MnO4−(aq) + 5(SO3)2−(aq) + H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + H2O(l)
I see on the left 16+30 O on the left, on the right 40 0 on the right (not cournting the water O), so finally then
4MnO4−(aq) + 5(SO3)2−(aq) + H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + 6H2O(l)
that balances the O
then finally, the H
4MnO4−(aq) + 5(SO3)2−(aq) + 6H+(aq) → 2Mn2+(aq) + 5(SO4)2−(aq) + 6H2O(l)
Now recheck it all.
For some reason its still not working
Forget it i figured it out
2MnO4{-}(aq)+5 SO3{2-}(aq)+6 H{+}(aq)→2 Mn{2+}(aq) + 5 SO4{2-}(aq) + 3 H2O(l)
2 I{-}(aq) + 4 H{+}(aq) + 2 NO2{-}(aq) → 2 NO(g) + 2 H2O(l) + I2(s)
To balance the equation, we need to ensure that the number of each type of atom on both sides of the equation is equal.
Let's break down the equation step by step:
1. Start with the unbalanced equation:
MnO4^-(aq) + SO32^-(aq) + H+(aq) → Mn2+(aq) + SO42^-(aq) + H2O(l)
2. Begin by balancing the atoms other than oxygen and hydrogen. In this case, we have only one type of atom other than oxygen and hydrogen: sulfur (S). Count the number of sulfur atoms on both sides of the equation:
On the left side: 1 sulfur atom
On the right side: 1 sulfur atom
Since the number of sulfur atoms is already balanced, we move on to the next step.
3. Next, balance the oxygen atoms. To do this, we can add water molecules to balance the oxygen on the left side:
On the right side, there are already 4 oxygen atoms in SO42^-. Therefore, we need to add 4 water molecules (H2O) to the left side:
MnO4^-(aq) + SO32^-(aq) + H+(aq) → Mn2+(aq) + SO42^-(aq) + 4H2O(l)
4. Now, we have 8 hydrogen atoms on the left side and only 4 hydrogen atoms on the right side. To balance the hydrogen atoms, we add 8 hydrogen ions (H+) to the right side:
MnO4^-(aq) + SO32^-(aq) + H+(aq) → Mn2+(aq) + SO42^-(aq) + 4H2O(l) + 8H+(aq)
5. Finally, we balance the charge on both sides of the equation. On the left side, the overall charge is -1 (MnO4^-). On the right side, we have Mn2+ and SO42-, which have a total charge of +2. To balance the charges, we add 1 electron (e^-) to the left side. This will give us a total charge of -2 on the left side:
MnO4^-(aq) + SO32^-(aq) + H+(aq) + e^- → Mn2+(aq) + SO42^-(aq) + 4H2O(l) + 8H+(aq)
Now that the equation is balanced in terms of atoms and charges, we can simplify it by canceling out the common species present on both sides of the arrow:
1MnO4^-(aq) + 2SO32^-(aq) + 8H+(aq) + 5e^- → 1Mn2+(aq) + 2SO42^-(aq) + 4H2O(l)
Therefore, the balanced equation can be written as:
1MnO4^-(aq) + 2SO32^-(aq) + 8H+(aq) + 5e^- → 1Mn2+(aq) + 2SO42^-(aq) + 4H2O(l)
The integer representation of the balanced equation is 128524580.