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
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
1 answer
1 answer