Asked by Nicole
                how do you balance a redox equation like:
Bi(NO3)3 + Al + NaOH = Bi + NH3 + NaAlO2 ??
            
        Bi(NO3)3 + Al + NaOH = Bi + NH3 + NaAlO2 ??
Answers
                    Answered by
            Dr Russ
            
    Bi(NO3)3 + Al + NaOH = Bi + NH3 + NaAlO2
If you want to balance an equation by inspection, then a good tip is to leave atoms alone and work with the molecules first.
An alternative method depends on how good your basic algebra is?
Bi(NO3)3 + Al + NaOH = Bi + NH3 + NaAlO2
lets say the missing numbers are as below
aBi(NO3)3 + bAl + cNaOH = Bi + dNH3 + eNaAlO2
(I have assumed that the number in front of Bi=1)
by balancing the atoms we can come up with a series of simultaneous equations
Bi a=1
N 3a=d
O 9a+c=2e
Al b=e
Na c=e
so we have five equations and five unknowns which we can solve.
a=1 so d=3 which is easy!
substitute value for a into the third equation
9+c=2e and we know c=e from the fifth equation
9+e=2e
hence e=9, c=9, and b=9
replace the symbols for numbers in the original
Bi(NO3)3 + 9Al + 9NaOH = Bi + 3NH3 + 9NaAlO2
and this method works EVERY time!!
    
If you want to balance an equation by inspection, then a good tip is to leave atoms alone and work with the molecules first.
An alternative method depends on how good your basic algebra is?
Bi(NO3)3 + Al + NaOH = Bi + NH3 + NaAlO2
lets say the missing numbers are as below
aBi(NO3)3 + bAl + cNaOH = Bi + dNH3 + eNaAlO2
(I have assumed that the number in front of Bi=1)
by balancing the atoms we can come up with a series of simultaneous equations
Bi a=1
N 3a=d
O 9a+c=2e
Al b=e
Na c=e
so we have five equations and five unknowns which we can solve.
a=1 so d=3 which is easy!
substitute value for a into the third equation
9+c=2e and we know c=e from the fifth equation
9+e=2e
hence e=9, c=9, and b=9
replace the symbols for numbers in the original
Bi(NO3)3 + 9Al + 9NaOH = Bi + 3NH3 + 9NaAlO2
and this method works EVERY time!!
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