MnO4−(aq) + SO32−(aq) + H+(aq) → Mn2+(aq) + SO42−(aq) + H2O(l)
in the form
a*MnO4−(aq) + b*SO32−(aq) + c*H+(aq) → d*Mn2+(aq) + e*SO42−(aq) + f*H2O(l)
Start by counting how many of each are on each side. I'm going to choose to use the convention [element (left, right)]
b=e; (Balancing the sulfur, S)
a=d; (Balancing the Mn)
c=2*f; (Balancing the hydrogen)
3*b + 4*a = 4*e + f (Balancing the oxygen)
So let a=e = 1; b=1; f=3
(3*1 + 4*1 = 4*1 + 3)
So the equation is
MnO4−(aq) + SO32−(aq) + 6H+(aq) → Mn2+(aq) + SO42−(aq) + 3H2O(l)
Al(s) + NO3−(aq) + OH−(aq) + H2O → Al(OH)4−(aq) + NH3(g)
of the form
a*Al(s) + b*NO3−(aq) + c*OH−(aq) + d*H2O → e*Al(OH)4−(aq) + f*NH3(g)
a=e (balancing the aluminum)
b=f (balancing the nitrogen)
3*b + c + d = 4*e (balancing the oxygen)
c + 2*d = 4*e + 3*f (Balancing the hydrogen)
You should be able to experiment with numbers and balance this out
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