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anonymous
 5 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.
anonymous
 5 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.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0dont you think that eqn is wrong because in reactant side there are no hydrogen atoms?????????????/

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I looked at the question again...thats what it says! Thats why I am so confused!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you have to balance it using ion electron method

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i haven't learned that yet....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but what I'm saying is that I don't know how to do it, can you explain?

JFraser
 5 years ago
Best ResponseYou've already chosen the best response.0I 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 NO31) to a +3 (in NH3). Split these 2 up into halfreactions\[Al{^0} \rightarrow Al{^+}{^3}\]\[NO{_3}{^}{^1} \rightarrow NH{_3}\] Each halfreaction 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 halfreactions are balanced for mass and charge. We need to add them back together again, but the electrons have to cancel. Multiply the aluminum halfreaction by 8, and multiply the nitrogen halfreaction 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 halfreactions back together and cancel what you can.\[8 Al + 24e{^}{^1} + 72H{^+} + 72OH{^}{^1} + 8NO{_3}{^}{^1} \rightarrow 8Al{^+}{^3} + 24e{^}{^1} + 8NH{_3} + 42H{_2}O + 72OH{^}{^1}\]
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