# Balance the following chemical equation a(Cr_2O_7)^(2-) + b(H)^(+) c(H_2SO_3) --&gt; d(Cr)^(3+) + e(H_2O)+ f(HSO_4)^(-)

jerichorayel | College Teacher | (Level 2) Senior Educator

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`Cr2O7^(2-) + H^(+) + H2SO3 -> Cr^(3+) + H2O + HSO4^(-)`

Looking at the equation, the first thing that you can balance is the number of Cr. By putting 2 beside `Cr^(3+)`  you can balance Cr. By the way, S and H are also balanced. So we try to work on the number of O's. (Remember that this is not a simple equation to balance.)

`Cr2O7^(2-) + H^(+) + H2SO3 -> 2Cr^(3+) + H2O + HSO4^(-)`

By putting 2's on H2O and HSO4-, O will be balanced but H and S will be affected.

`Cr2O7^(2-) + H^(+) + H2SO3 -> 2Cr^(3+) + 2H2O + 2HSO4^(-)`

If we try to resolve it by adding 2 beside H2SO4, S will be balanced but O's and H' will be changed.

`Cr2O7^(2-) + H^(+) + 2H2SO3 -> 2Cr^(3+) + 2H2O + 2HSO4^(-)`

Now we try to balance O gain by changing the coefficient of HSO4- from 2 to 3. By doing that we should change H2SO4 as well to balance S.

`Cr2O7^(2-) + H^(+) + 3H2SO3 -> 2Cr^(3+) + 2H2O + 3HSO4^(-)`

By changing the coefficient of H2O from 2 to 4, the O will be balanced.

`Cr2O7^(2-) + H^(+) + 3H2SO3 -> 2Cr^(3+) + 4H2O + 3HSO4^(-)`

FINALLY, by counting the H's, we can resolve it by putting 5 beside H+.

`Cr2O7^(2-) + 5H^(+) + 3H2SO3 -> 2Cr^(3+) + 4H2O + 3HSO4^(-)`

By counting the number of atoms in both sides we have:

Cr -> 2

O -> 16

H -> 11

S -> 3

So our equation is balanced.