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Balance the following equation xH_(2)S+yHNO_(3) to zNO +omegaS+H_(2)O The coefficients omega, x ,y, z" are" (1) x=3, y=2, z=2, omega=3 (2) x=2, y=2, z=3, omega=3 (3) x=3, y=3, z=2, omega=3 (4) x=3, y=2, z=3, omega=3 |
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Answer» Solution :Step -(i) The two half reaction of the equation are: Oxidation : `H_(2)S to S` REDUCTION `NO_(3)^(-) to NO` Step (II) A. Balancing of the oxidation half reaction (a) It is balanced by adding `2H^+` on the R.H.S. `H_(2)S to S+2H^(+)` (b) For equalising charge on both sides, add two ELECTRONS on the R.H.S. of the above equation. `H_(2)S to S+2H^(+) +2e^(-) ......(i)` B. Balancing of the reduction half reaction. `NO_(3)^(-) to NO` (a) For balancing oxygen atoms, add `2H_2O` to R.H.S. `NO_(3)^(-) to NO+2H_(2)O` (b) For balancing H atoms, add `4H^+` to L.H.S. of the above equation `NO_(3)^(-)+4H^(+) to NO+2H_(2)O` (C) For balancing charge on both sides of the above equation, add `3e^-` to L.H.S. `NO_(3)^(-) + 4H^+ + 3e^- to NO+2H_(2)O .........(ii)` From (i) & (ii) we get, `H_(2)S to S+2H^(+) +2e^(-) ] xx 3` `NO_(3)^(-)+4H^(+)+3e^(-) to NO+2H_(2)O] xx 2` `2H_(2)S+8H^(+)+2NO_(3)^(-) to 3S+2NO+4H_(2)O+6H^(+)` Step (III) On cancelling the common terms, the final balanced equation becomes `3H_(2)S+2HNO_(3) to 2NO+3S+4H_(2)O` |
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