# CHEMISTRY

Write a balanced equation for the following reduction-oxidation reaction.

SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)

can someone help me step by step??

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2. SO3(2-) + MnO4(-) --- SO4(2–) + Mn(2+)
first deal with S, Mn atoms
there must be the same number on each side

nSO3(2-) + mMnO4(-) --- nSO4(2–) + mMn(2+)
now try 2jH(+) on the left and jH2O on the right
nSO3(2-) + mMnO4(-) +2jH(+) --- nSO4(2–) + mMn(2+) + jH2O
O atoms
3 n + 4 m = 4 n + j
or
n+j = 4 m
+ charges
-2 n - 1 m + 2 j = -2 n + 2 m
so from the second one
3 m = 2 j
j = (3/2) m
from the first one
n + (3/2) m = 4 m
or
n = (5/2) m
Now go back using m = 2 to make the fraction go away
then j = 3
and n = 5
5 SO3(-2) + 2 MnO4(-1) + 6 H(+1) --> 3 H2O + 5 SO4(-2) + 2 Mn(+2)

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posted by Damon

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