In the equation xP+ HNO_(3) to HPO_(3)+yNO +H_(2)O

1.	In the equation xP+ HNO_(3) to HPO_(3)+yNO +H_(2)O
Answer» x=5,y=5 x=5, y=3 x=3, y=5 x=3, y=3 Solution :Step -I. `P+HNO_(3) to HPO_(3)+NO+H_(2)O` Step -II. Writing the oxidation number of the ELEMENTS `P+HNO_(3) to HPO_(3)+NO+H_(2)O` `0+ +5 ""+5 ""+2` Thus here P is oxidized (0 to +5) and N in `HNO_(3)` is reduced (+5 to +2). Writing seperate EQUATION for the reducing agent (P) and oxidising agent `("N in "HNO_(3))`. Oxidation `P^(0) to P^(+5) uparrow 5` Reduction `N^(+5) to N^(+2) downarrow 3` Step -III. Adding electrons according to change in oxidation number. `P^(0) to P^(+5) +5e^(-) .......(i)` `N^(+5) +3e^(-) to N^(+2) .......(ii)` Step -IV. Multiply equation (i) by (3) and (ii) by 5. `3P^(0) to 3P^(+5) +15e^(-)..........(iii)` `5N^(+5)+15e^(-) to 5N^(+2) .......(iv)` Adding (iii) and (iv) `3P^(0)+5N^(+5) to 3P^(+5) +5N^(+2)` Therefore, the required reaction can be WRITTEN as `+5HNO_(3) to 3HPO_(3)+5NO` Hence the required balanced equation can be REPRESENTED as below `3P+5HNO_(3) to 3HPO_(3)+5NO+H_(2)O`

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