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 got 2525 and its going in as incorrect

Now you get to balance this equation (answer in the same way as in the problem above):

Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)

i go 2121 but its not correct

Start from scratch and separate into half-reactions

MnO4-(aq) -----> Mn(^2+)(aq)
SO3(^2-)(aq) ----> SO4(^2-)(aq)

Balance O by adding H2O as needed:
MnO4-(aq) -----> Mn(^2+)(aq) + 4H2O(l)
SO3(^2-)(aq) + H2O(l) ----> SO4(^2-)(aq)

balance H by adding H+:
MnO4-(aq) + 8H+(aq)-----> Mn(^2+)(aq) + 4H2O(l)
SO3(^2-)(aq) + H2O(l) ----> SO4(^2-)(aq) + 2H+(aq

Balance charge by adding e-:
MnO4-(aq) + 8H+(aq) + 5e- -----> Mn(^2+)(aq) + 4H2O(l)
SO3(^2-)(aq) + H2O(l) ----> SO4(^2-)(aq) + 2H+(aq) + 2e-

Equalize e- in the two equations
2MnO4-(aq) + 16H+(aq) + 10e- -----> 2Mn(^2+)(aq) + 8H2O(l)
5SO3(^2-)(aq) + 5H2O(l) ----> 5SO4(^2-)(aq) + 10H+(aq) + 10e-

Add the equations:
2MnO4-(aq) + 16H+(aq) + 10e- + 5SO3(^2-)(aq) + 5H2O(l) ----> 2Mn(^2+)(aq) + 8H2O(l) + 5SO4(^2-)(aq) + 10H+(aq) + 10e-

Cancel like species appearing on both sides
2MnO4-(aq) + 6H+(aq) + 5SO3(^2-)(aq) ----> 2Mn(^2+)(aq) + 3H2O(l) + 5SO4(^2-)(aq)

Ans = 2525

Al(s) → Al(OH)4−(aq)
NO3−(aq) --->NH3(g)

Al(s) + 4H2O(l) → Al(OH)4−(aq)
NO3−(aq) --->NH3(g) + 3H2O(l)

Al(s) + 4H2O(l) → Al(OH)4−(aq) + 4H+(aq)
NO3−(aq) + 9H+(aq) --->NH3(g) + 3H2O(l)

Al(s) + 4H2O(l) → Al(OH)4−(aq) + 4H+(aq) + 4e-
NO3−(aq) + 9H+(aq) + 8e- --->NH3(g) + 3H2O(l)

2Al(s) + 8H2O(l) → 2Al(OH)4−(aq) + 8H+(aq) + 8e-
NO3−(aq) + 9H+(aq) + 8e- --->NH3(g) + 3H2O(l)

2Al(s) + 8H2O(l) + NO3−(aq) + 9H+(aq) + 8e- ----> 2Al(OH)4−(aq) + 8H+(aq) + 8e- + NH3(g) + 3H2O(l)

2Al(s) + 5H2O(l) + NO3−(aq) + H+(aq)----> 2Al(OH)4−(aq) + NH3(g)

Add OH- to neutralize H+

2Al(s) + 6H2O(l) + NO3−(aq) ----> 2Al(OH)4−(aq) + NH3(g) + OH-(aq)

Ans = 2121

To balance the given chemical equation:

Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)

We need to ensure that the number of each type of atom is the same on both sides of the equation. Let's go step by step:

1. Balance the aluminum (Al) atoms:
There is only one Al atom on both sides of the equation, so it is already balanced.

2. Balance the nitrogen (N) atoms:
There is one N atom in the NO3− ion on the left side of the equation, and no N atom on the right side. Add a coefficient of 2 in front of NH3(g) to balance the N atoms.

Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + 2NH3(g)

3. Balance the hydrogen (H) atoms:
There are 2 H atoms in the Al(OH)4− ion on the right side of the equation. We have 1 H atom in the OH− ion and 2 H atoms in the H2O molecule on the left side of the equation. To balance the H atoms, add a coefficient of 6 in front of H2O.

Al(s) + NO3−(aq) + OH−(aq) + 6H2O → Al(OH)4−(aq) + 2NH3(g)

4. Balance the oxygen (O) atoms:
There are 6 O atoms in the Al(OH)4− ion and 2 O atoms in the NO3− ion on the right side of the equation. We have 1 O atom in the OH− ion and 6 O atoms in the 6H2O molecules on the left side of the equation. To balance the O atoms, add a coefficient of 4 in front of the OH− ion.

Al(s) + NO3−(aq) + 4OH−(aq) + 6H2O → Al(OH)4−(aq) + 2NH3(g)

Now, count the coefficients of each compound and write them as a single number:
1 0106 + 1104 + 4016 + 6010 = 10104216

So, the balanced equation can be represented as 10104216.