Asked by Nikki
Balance the equation:
NO2- (aq) + Al (s) --> NH3 (g) + AlO2- (aq) (basic solution)
** please note the "2-" on the NO2- and AlO2- are not subscripts, they are the charges.
I missed a few classes in Chemistry and find myself lost. I have tried to learn these equations with the steps but I am still not able to solve them. Please show work when answering. Thank you so much!
NO2- (aq) + Al (s) --> NH3 (g) + AlO2- (aq) (basic solution)
** please note the "2-" on the NO2- and AlO2- are not subscripts, they are the charges.
I missed a few classes in Chemistry and find myself lost. I have tried to learn these equations with the steps but I am still not able to solve them. Please show work when answering. Thank you so much!
Answers
Answered by
DrBob222
I don't think that equation is correct. For AlO^2- to be correct means Al must be zero and there is no redox when Al(s) goes to AlO^2- since Al is zero in that anion. Same for NO^2-. I might believe [AlO2]^- and [NO2]^-
1. Al ==> [AlO2]^-
2. Assign oxidation states and add electrons to balance the change.
Al is zero on left and +3 on the right.
Al ==> [AlO2]- + 3e
3. Count the charges on each side (zero on left and -4 on the right; add OH^- to balance the charge.
Al + 4OH^- ==> [AlO2]^- + 3e
4. Now add H2O to the appropriate side to balance the H.
Al + 4OH^- ==> [AlO2]^- + 3e + 2H2O
The NO2^- is done the same way.
1. Al ==> [AlO2]^-
2. Assign oxidation states and add electrons to balance the change.
Al is zero on left and +3 on the right.
Al ==> [AlO2]- + 3e
3. Count the charges on each side (zero on left and -4 on the right; add OH^- to balance the charge.
Al + 4OH^- ==> [AlO2]^- + 3e
4. Now add H2O to the appropriate side to balance the H.
Al + 4OH^- ==> [AlO2]^- + 3e + 2H2O
The NO2^- is done the same way.
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