Question

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.

Answer 1

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

Answer 2

r u there?

Answer 3

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

Answer 4

so what it says

Answer 5

you have to balance it using ion electron method

Answer 6

i haven't learned that yet....

Answer 7

so, you have to

Answer 8

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

Answer 9

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.\[8 Al + 24e{^-}{^1} + 72H{^+} + 72OH{^-}{^1} + 8NO{_3}{^-}{^1} \rightarrow 8Al{^+}{^3} + 24e{^-}{^1} + 8NH{_3} + 42H{_2}O + 72OH{^-}{^1}\]