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(i) K(2)Cr(2)O(7)+KI+H(2)SO(4)to K(2)SO(...

(i) `K_(2)Cr_(2)O_(7)+KI+H_(2)SO_(4)to K_(2)SO_(4)+Cr_(2)(SO_4)_3+I_(2)+H_(2)O`.

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Equalise the incrase / decrease in O N by multiplying I species by 1
`K_(2)Cr_(2)O_(7)+3KI+H_(2)SO_(4)to K_(2)SO_(4)+Cr_(2)(SO_4)_3+3I_(2)+H_2O`.
Balance all other atoms except H and O
`K_2Cr_(2)O_7 +6KI+7H_(2)SO_(2)to 4K_(2)SO_(4)+Cr_(2)(SO_4)_3+3I_2+H_2O`
Balance O atom by adding `H_2O` on the side falling short of oxygen
`K_2Cr_(2)O_(7)+6KI+7H_(2)SO_(4)to 4K_(2)SO_(4)+Cr_(2)(SO_4)_3+3I_(2)+H_(2)+6H_(2)O`
So the balanced equation is
`K_(2)Cr_(2)O_(7)+6KI+7H_2SO_(4)to 4K_2SO_(4)+Cr_(2)(SO_4)_3+3I_(2)+7H_2 O` .
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