Balance the net equtation fro th reaction of potassium dichromate (VI), `K_(2) Cr_(2) O_(7)`, with sodium sulphite, `Na_(2)SO_(3)`, in an acid solution to give chromium (III) ion and and sulphate ion.
Strategy : Follow the seven -step proceduce , one step at a time.

Explanation

Step 1 write the skeleton equation for the given reaction
`Cr_(2)O_(7)^(2-)(aq)+SO_(3)^(2-)(aq)rarrCr^(3+)(aq)+SO_(4)^(2-)(aq)`
Step 2 find out the element which undergo a change in oxidiatoin number (O.N)

here O.N of Cr decrease from +6 `Cr_(2)O_(7)^(2-)` to +3 in `Cr^(3+)` while that of s increases from +4 in `SO_(3)^(2+) to +6 in SO_(4)^(2-)`
step 3 find out the total increase and decrease in O.N
since there are two Cr atoms on L.H.S and only one on R.H.S therefore multiply `Cr^(3+)` onR.H.S of eq (i) by 2 and thus the total decrease in O.HN of Cr is `2xx3=6`
step 4 balance increase / decrease in O.N
since the toal increase in O.N is 2 and decrease is 6 therefore multiply `SO_(3)^(2-)` on L.H.S and `SO_(4)^(2-)` on R.H.S of Eq (ii) 3 combining steps 2 and 3 we have
step 5 balance all atoms than H and O not needed since both cr and s atoms are already balanced
step 6 balance o atoms by adding `H_(2)O` moleculed
since there are seven O atoms in `Cr_(2)O_(7)^(2-)` and nine in `3So_(3)^(2-)` on L.H.S and only 12 on the R.H.S of eq (iii) therefore add 4 molecules of `H_(2)O` to R.H.S of eq (iii) we have
`Cr_(2)O_(7)^(2-) (aq)+3SO_(3)^(2-)(aq)rarr2cr^(3+)(aq)+3SO_(4)^(2-)(aq)+4H_(2)O(l)` step 7 balance H toms by adding `H^(+)` ions since the reaction occurs in the acidic medium since there are 8H on R.H.s and none o the L.H.S therefore add `8H^(+)` to the L.H.S of Eq (iv) we have
`Cr_(2)O_(7)^(2-)+3SO_(3)^(2-)(aq)+8H^(+)(aq)rarr2Cr^(3+)r(aq)+3SO_(4)^(2-)(aq)+4H_(2)O(l)`
thus equ (v) represent the correct balanced equation