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Some Basic Concepts Of Chemistry

Question
CBSEENCH11005062

Balance the following skeleton equation by hit and trial method:
Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 plus space NH subscript 3 plus straight H subscript 2 straight O space rightwards arrow space space space Fe left parenthesis OH right parenthesis subscript 3 plus left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4

Solution
(i) The skeleton equation is
Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 plus space NH subscript 3 plus straight H subscript 2 straight O space rightwards arrow space space space Fe left parenthesis OH right parenthesis subscript 3 plus left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
(ii) To balance Fe atoms, multiply Fe(OH)by 2.
Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 plus NH subscript 3 plus straight H subscript 2 straight O space rightwards arrow space space 2 Fe left parenthesis OH right parenthesis subscript 3 plus space left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
(iii) To balance S atoms, multiply (NH4)2SOby 3.
Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 plus NH subscript 3 plus straight H subscript 2 straight O space space rightwards arrow space space space space 2 Fe left parenthesis OH right parenthesis subscript 3 plus 3 left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
(iv) To balance N atoms, multiply NHby 6.
Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 plus 6 NH subscript 3 plus straight H subscript 2 straight O space space rightwards arrow space space space space 2 Fe left parenthesis OH right parenthesis subscript 3 plus 3 left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
(v) To balance H and O atoms, multiplyng H2O by 6, we get.
Fe subscript 2 left parenthesis SO subscript 4 right parenthesis subscript 3 plus 6 NH subscript 3 plus 6 straight H subscript 2 straight O space rightwards arrow space space 2 Fe left parenthesis OH right parenthesis subscript 3 plus 3 left parenthesis NH subscript 4 right parenthesis subscript 2 SO subscript 4
This is the balanced equation.