## anonymous 4 years ago Balance the following equation in basic solution: Al + NO3 --> Al(OH)4 + NH3 I need to find the coefficients for Al, NO3, OH, H2O, Al(OH)4, and NH3.

1. anonymous

dont you think that eqn is wrong because in reactant side there are no hydrogen atoms?????????????/

2. anonymous

r u there?

3. anonymous

I looked at the question again...thats what it says! Thats why I am so confused!

4. anonymous

so what it says

5. anonymous

6. anonymous

you have to balance it using ion electron method

7. anonymous

i haven't learned that yet....

8. anonymous

so, you have to

9. anonymous

but what I'm saying is that I don't know how to do it, can you explain?

10. JFraser

I think you've got a typo or two, so I'll make my best guess. Find the oxidation states of the aluminum and nitrogen atoms in the unbalanced reaction. The aluminum goes from a zero (Al by itself) to a +3 (Al+3 inside the ion Al(OH)4^-1). The nitrogen goes from a +5 (in the NO3-1) to a +3 (in NH3). Split these 2 up into half-reactions$Al{^0} \rightarrow Al{^+}{^3}$$NO{_3}{^-}{^1} \rightarrow NH{_3}$ Each half-reaction has to be balanced for MASS first, then CHARGE. The aluminum is easiest, all it needs is 3 electrons to balance charge $Al{^0} \rightarrow Al{^+}{^3} + 3e{^-}{^1}$ The aluminum has been oxidized (lost electrons). The nitrogen is a little more complicated. Balance the oxygens next, using water$NO{_3}{^-}{^1} \rightarrow NH{_3} + 3H{_2}O$ Balance the hydrogens next, using H+ ions $9H{^+} + NO{_3}{^-}{^1} \rightarrow NH{_3} + 3H{_2}O$ Since the solution has to be basic, there can't be any random H+ ions floating around, so every H+ has to be cancelled by a hydroxide$9H{^+} + 9OH{^-}{^1} + NO{_3}{^-}{^1} \rightarrow NH{_3} + 3H{_2}O + 9OH{^-}{^1}$ This balances for mass, but not charge. The total charge of the reactants is -1, but the total charge of the products is -9. EIGHT electrons have to be added to the reactants$8e{^-}{^1} + 9H{^+} + 9OH{^-}{^1} + NO{_3}{^-}{^1} \rightarrow NH{_3} + 3H{_2}O + 9OH{^-}{^1}$ Now both half-reactions are balanced for mass and charge. We need to add them back together again, but the electrons have to cancel. Multiply the aluminum half-reaction by 8, and multiply the nitrogen half-reaction by 3 so that 24 electrons cancel 24 electrons$8Al{^0} \rightarrow 8Al{^+}{^3} + 24e{^-}{^1}$$24e{^-}{^1} + 72H{^+} + 72OH{^-}{^1} + 8NO{_3}{^-}{^1} \rightarrow 8NH{_3} + 24H{_2}O + 72OH{^-}{^1}$ Add the half-reactions back together and cancel what you can.$8 Al + 24e{^-}{^1} + 72H{^+} + 72OH{^-}{^1} + 8NO{_3}{^-}{^1} \rightarrow 8Al{^+}{^3} + 24e{^-}{^1} + 8NH{_3} + 42H{_2}O + 72OH{^-}{^1}$